#include <bits/stdc++.h>
using namespace std;
int dx[8] = {0, 0, 1, -1, 1, 1, -1, -1};
int dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
template <typename T>
inline T BigMod(T A, T B) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % 1000000007;
    A = (A * A) % 1000000007;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T InvMod(T A, T M = 1000000007) {
  return BigMod(A, M - 2);
}
string tostring(long long int res) {
  string curstr = "";
  while (res != 0) {
    long long int temp = (res % 10);
    curstr += ((char)temp + '0');
    res /= 10;
  }
  reverse(curstr.begin(), curstr.end());
  return curstr;
}
long long int toint(string ss) {
  long long int inss = 0;
  for (int i = 0; i < ss.size(); i++) {
    inss = (inss * 10) + ((int)(ss[i] - '0'));
  }
  return inss;
}
int ar[1503], n, m;
int main() {
  while (cin >> n) {
    for (int i = 0; i < n; i++) cin >> ar[i];
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < i; j++) {
        if (ar[j] > ar[i]) cnt++;
      }
    }
    scanf("%d", &m);
    int u, v, track = (cnt & 1);
    while (m--) {
      scanf("%d %d", &u, &v);
      int ind = (v - u) + 1;
      int pairs = (ind * (ind - 1)) / 2;
      if (pairs & 1) track = 1 - track;
      track ? printf("odd\n") : printf("even\n");
    }
  }
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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 dx[8] = {0, 0, 1, -1, 1, 1, -1, -1};
int dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
template <typename T>
inline T BigMod(T A, T B) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % 1000000007;
    A = (A * A) % 1000000007;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T InvMod(T A, T M = 1000000007) {
  return BigMod(A, M - 2);
}
string tostring(long long int res) {
  string curstr = "";
  while (res != 0) {
    long long int temp = (res % 10);
    curstr += ((char)temp + '0');
    res /= 10;
  }
  reverse(curstr.begin(), curstr.end());
  return curstr;
}
long long int toint(string ss) {
  long long int inss = 0;
  for (int i = 0; i < ss.size(); i++) {
    inss = (inss * 10) + ((int)(ss[i] - '0'));
  }
  return inss;
}
int ar[1503], n, m;
int main() {
  while (cin >> n) {
    for (int i = 0; i < n; i++) cin >> ar[i];
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < i; j++) {
        if (ar[j] > ar[i]) cnt++;
      }
    }
    scanf("%d", &m);
    int u, v, track = (cnt & 1);
    while (m--) {
      scanf("%d %d", &u, &v);
      int ind = (v - u) + 1;
      int pairs = (ind * (ind - 1)) / 2;
      if (pairs & 1) track = 1 - track;
      track ? printf("odd\n") : printf("even\n");
    }
  }
  return 0;
}

```