Codeforces Round #497 (Div. 2) ABC

A. Romaji

Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.

In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.

Help Vitya find out if a word $\mathrm{ss is\; Berlanese.}$

Input

The first line of the input contains the string $\mathrm{ss consisting\; of |s||s| (1\le |s|\le 1001\le |s|\le 100)\; lowercase\; Latin\; letters.}$

Output

Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".

You can print each letter in any case (upper or lower).

Examples

input

Copy

sumimasen

output

Copy

YES

input

Copy

ninja

output

Copy

YES

input

Copy

codeforces

output

Copy

NO

Note

In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.

In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.

辅音字母除n外后面是否都有元音字母

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
char str[110];
map<char,int> mp;
int main() {
    mp['e'] = mp['a'] = mp['o'] = mp['u'] = mp['i'] = 1;
    cin >> str;
    for(int i = 0; str[i]; i ++) {
        if(str[i] == 'n') continue;
        if(mp[str[i]] == 0 ) {
            if(mp[str[i+1]] != 1) {
                // printf("%c %c
",str[i],str[i+1] );
                return 0*printf("no
");
            }
        }
    }printf("yes
");
    return 0;
}

B. Turn the Rectangles

There are $\mathrm{nn rectangles\; in\; a\; row.\; You\; can\; either\; turn\; each\; rectangle\; by 9090 degrees\; or\; leave\; it\; as\; it\; is.\; If\; you\; turn\; a\; rectangle,\; its\; width\; will\; be\; height,\; and\; its\; height\; will\; be\; width.\; Notice\; that\; you\; can\; turn\; any\; number\; of\; rectangles,\; you\; also\; can\; turn\; all\; or\; none\; of\; them. You\; can\; not\; change\; the\; order\; of\; the\; rectangles.}$

Find out if there is a way to make the rectangles go in order of non-ascending height. In other words, after all the turns, a height of every rectangle has to be not greater than the height of the previous rectangle (if it is such).

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; number\; of\; rectangles.}$

Each of the next $\mathrm{nn lines\; contains\; two\; integers wiwi and hihi (1\le wi,hi\le 1091\le wi,hi\le 109) —\; the\; width\; and\; the\; height\; of\; the ii-th\; rectangle.}$

Output

Print "YES" (without quotes) if there is a way to make the rectangles go in order of non-ascending height, otherwise print "NO".

You can print each letter in any case (upper or lower).

Examples

input

Copy

3
3 4
4 6
3 5

output

Copy

YES

input

Copy

2
3 4
5 5

output

Copy

NO

Note

In the first test, you can rotate the second and the third rectangles so that the heights will be [4, 4, 3].

In the second test, there is no way the second rectangle will be not higher than the first one.

有一个位置有两个数字，通过适当的选择是否存在递减序列

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
int x[N], y[N];
int main() {
    int n;
    cin >> n;
    for(int i = 1; i <= n; i ++) cin >> x[i] >> y[i];
    int cnt = max(x[1],y[1]);
    for(int i = 2; i <= n; i ++) {
        if(cnt >= max(x[i],y[i])) {
            cnt = max(x[i],y[i]);
        } else if(cnt >= min(x[i],y[i])) {
            cnt = min(x[i], y[i]);
        } else return 0*printf("NO
");
    }
    printf("YES
");
    return 0;
}

C. Reorder the Array

You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.

For instance, if we are given an array $[10,20,30,40][10,20,30,40],\; we\; can\; permute\; it\; so\; that\; it\; becomes [20,40,10,30][20,40,10,30].\; Then\; on\; the\; first\; and\; the\; second\; positions\; the\; integers\; became\; larger\; (20>1020>10, 40>2040>20)\; and\; did\; not\; on\; the\; third\; and\; the\; fourth,\; so\; for\; this\; permutation,\; the\; number\; that\; Vasya\; wants\; to\; maximize\; equals 22.\; Read\; the\; note\; for\; the\; first\; example,\; there\; is\; one\; more\; demonstrative\; test\; case.$

Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; length\; of\; the\; array.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109) —\; the\; elements\; of\; the\; array.}$

Output

Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.

Examples

input

Copy

7
10 1 1 1 5 5 3

output

Copy

4

input

Copy

5
1 1 1 1 1

output

Copy

0

Note

In the first sample, one of the best permutations is $[1,5,5,3,10,1,1][1,5,5,3,10,1,1].\; On\; the\; positions\; from\; second\; to\; fifth\; the\; elements\; became\; larger,\; so\; the\; answer\; for\; this\; permutation\; is\; 4.$

In the second sample, there is no way to increase any element with a permutation, so the answer is 0.

最多存在多少个将大的数字放在小的数字上。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
int n, a[N];
int main() {
    cin >> n;
    for(int i = 0; i < n; i ++) cin >> a[i];
    sort(a,a+n);
    int tmp = n-2;
    int ans = 0;
    for(int i = n-1; i >= 0; i --) {
        while(a[tmp] >= a[i] && tmp >= 0) tmp--;
        if(tmp < 0) break;
        ans++;tmp--;

    }
    cout << ans << endl;
    return 0;
}