#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 1e10;
template <typename T>
void cmax(T &a, T b) {
  a = max(a, b);
}
template <typename T>
void cmin(T &a, T b) {
  a = min(a, b);
}
vector<long long> a;
long long bit[100005];
long long n;
long long sum(long long i) {
  long long s = 0;
  while (i > 0) {
    s += bit[i];
    i -= i & -i;
  }
  return s;
}
void add(long long i, long long x) {
  while (i <= n) {
    bit[i] += x;
    i += i & -i;
  }
}
long long solve() {
  long long ans = 0;
  for (long long j = 0; j < n; j++) {
    ans += j - sum(a[j]);
    add(a[j], 1);
  }
  return ans;
}
signed main() {
  cin >> n;
  a.resize(n);
  for (long long i = 0; i < (n); i++) cin >> a[i];
  long long inversion = solve();
  bool isEven = true;
  if (inversion % 2 == 1) isEven = false;
  long long m;
  cin >> m;
  while (m--) {
    long long l, r;
    cin >> l >> r;
    long long range = r - l + 1;
    if (range * (range - 1) / 2 % 2 == 0) {
    } else {
      if (isEven)
        isEven = false;
      else
        isEven = true;
    }
    cout << (isEven ? "even" : "odd") << endl;
  }
  return 0;
}

### Prompt

Generate a cpp solution to 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;
const long long MOD = 1e9 + 7;
const long long INF = 1e10;
template <typename T>
void cmax(T &a, T b) {
  a = max(a, b);
}
template <typename T>
void cmin(T &a, T b) {
  a = min(a, b);
}
vector<long long> a;
long long bit[100005];
long long n;
long long sum(long long i) {
  long long s = 0;
  while (i > 0) {
    s += bit[i];
    i -= i & -i;
  }
  return s;
}
void add(long long i, long long x) {
  while (i <= n) {
    bit[i] += x;
    i += i & -i;
  }
}
long long solve() {
  long long ans = 0;
  for (long long j = 0; j < n; j++) {
    ans += j - sum(a[j]);
    add(a[j], 1);
  }
  return ans;
}
signed main() {
  cin >> n;
  a.resize(n);
  for (long long i = 0; i < (n); i++) cin >> a[i];
  long long inversion = solve();
  bool isEven = true;
  if (inversion % 2 == 1) isEven = false;
  long long m;
  cin >> m;
  while (m--) {
    long long l, r;
    cin >> l >> r;
    long long range = r - l + 1;
    if (range * (range - 1) / 2 % 2 == 0) {
    } else {
      if (isEven)
        isEven = false;
      else
        isEven = true;
    }
    cout << (isEven ? "even" : "odd") << endl;
  }
  return 0;
}

```