codeforces 814A An abandoned sentiment from past

A. An abandoned sentiment from past

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.

To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.

Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.

If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in awith an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.

Input

The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and brespectively.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.

The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.

Input guarantees that apart from 0, no integer occurs in a and b more than once in total.

Output

Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.

Examples

input

4 2
11 0 0 14
5 4

output

Yes

input

6 1
2 3 0 8 9 10
5

output

No

input

4 1
8 94 0 4
89

output

Yes

input

7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7

output

Yes

Note

In the first sample:

• Sequence a is 11, 0, 0, 14.
• Two of the elements are lost, and the candidates in b are 5 and 4.
• There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".

In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.

题意：给你n个数构成的序列a和k个数构成的序列b没有重复的非0数字。序列a中有k个0，现在要你用序列b的数字替换序列a中的0.问你如果可以构成非递增序列，输出Yes，否则输出No。

分析：由于每个数字都不一样，因此要想只能构成递增序列，就是k=1的情况且将b[1]替换a的0以后a是递增情况下输出No，其他情况都输出Yes。

AC代码：

#include<bits/stdc++.h>
using namespace std;

int a[205],b[205];
int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    for(int i=1;i<=k;i++){
        cin>>b[i];
    }
    if(k==1){
        for(int i=1;i<=n;i++){
            if(a[i]==0) {a[i]=b[1];break;}
        }
        int p=0;
        for(int i=1;i<n;i++){
            if(a[i+1]<a[i]){
                p=1;
                break;
            }
        }
        if(p==0) cout<<"No"<<endl;
        else cout<<"Yes"<<endl;
    }
    else cout<<"Yes"<<endl;
    return 0;
}