codeforces 911D

D. Inversion Counting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A permutation of size n is an array of size n such that each integer from 1 to n occurs exactly once in this array. An inversion in a permutation p is a pair of indices (i, j) such that i > j and a_i < a_j. For example, a permutation [4, 1, 3, 2] contains 4 inversions: (2, 1), (3, 1), (4, 1), (4, 3).

You are given a permutation a of size n and m queries to it. Each query is represented by two indices l and r denoting that you have to reverse the segment [l, r] of the permutation. For example, if a = [1, 2, 3, 4] and a query l = 2, r = 4 is applied, then the resulting permutation is [1, 4, 3, 2].

After each query you have to determine whether the number of inversions is odd or even.

Input

The first line contains one integer n (1 ≤ n ≤ 1500) — the size of the permutation.

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ n) — the elements of the permutation. These integers are pairwise distinct.

The third line contains one integer m (1 ≤ m ≤ 2·10⁵) — the number of queries to process.

Then m lines follow, i-th line containing two integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) denoting that i-th query is to reverse a segment [l_i, r_i] of the permutation. All queries are performed one after another.

Output

Print m lines. i-th of them must be equal to odd if the number of inversions in the permutation after i-th query is odd, and even otherwise.

Examples

Input

3
1 2 3
2
1 2
2 3

Output

odd
even

Input

4
1 2 4 3
4
1 1
1 4
1 4
2 3

Output

odd
odd
odd
even

Note

The first example:

1. after the first query a = [2, 1, 3], inversion: (2, 1);
2. after the second query a = [2, 3, 1], inversions: (3, 1), (3, 2).

The second example:

1. a = [1, 2, 4, 3], inversion: (4, 3);
2. a = [3, 4, 2, 1], inversions: (3, 1), (4, 1), (3, 2), (4, 2), (4, 3);
3. a = [1, 2, 4, 3], inversion: (4, 3);
4. a = [1, 4, 2, 3], inversions: (3, 2), (4, 2).

（这一题思路很巧妙

题意：给n个不相同的数，每次询问都会将l到r的数翻转，判断每次翻转后所有数的逆序数和的奇偶性。

解题思路：因为每段序列的逆序数和最大的情况是该序列从大到小排序，此时逆序数和为max=（r-l+1)*(r-l)/2 。那么设序列的逆序数和为m，则其翻转后为max-m。若max为偶数，则m与max-m的奇偶性一致，也就是说翻转无影响。而若max为奇数，则m与max-m的奇偶性不同，则翻转会改变答案。

也就是说，只需要先求出整个序列的逆序数和的奇偶性，再根据每次翻转，判断是否会改变答案的奇偶性就行了。

ac代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int maxn = 1500+10;
int nu[maxn];
int cnt=0;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
    scanf("%d",&nu[i]);
    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<i;++j)
        {
            if(nu[j]>nu[i])
            {
                cnt++;
            }
        }
    }
    cnt%=2;
    int m;
    scanf("%d",&m);
    int l,r;
    for(int i=1;i<=m;++i)
    {
        scanf("%d%d",&l,&r);
        if((r-l+1)*(r-l)/2%2==1) cnt^=1;
        if(cnt) printf("odd
");
        else printf("even
");
    }
    return 0;
}