#include <bits/stdc++.h>
using namespace std;
int n, a[1505], m;
bool p;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++)
    for (int j = i + 1; j <= n; j++)
      if (a[i] > a[j]) p = !p;
  cin >> m;
  while (m--) {
    int l, r;
    cin >> l >> r;
    if ((r - l + 1) % 4 && (r - l) % 4) p = !p;
    if (!p)
      cout << "even\n";
    else
      cout << "odd\n";
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 ai < aj. 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, ..., an (1 ≤ ai ≤ 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 li, ri (1 ≤ li ≤ ri ≤ n) denoting that i-th query is to reverse a segment [li, ri] 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). 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, a[1505], m;
bool p;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++)
    for (int j = i + 1; j <= n; j++)
      if (a[i] > a[j]) p = !p;
  cin >> m;
  while (m--) {
    int l, r;
    cin >> l >> r;
    if ((r - l + 1) % 4 && (r - l) % 4) p = !p;
    if (!p)
      cout << "even\n";
    else
      cout << "odd\n";
  }
}

```