#include <bits/stdc++.h>
using namespace std;
int arr[2 * 100500];
int p = 0, ne = 0;
int n, m;
int lesser = min(ne, p);
int a, b;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  cin >> m;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == -1) {
      p += 1;
    }
  }
  ne = abs(p - n);
  lesser = min(ne, p);
  while (m > 0) {
    cin >> a;
    cin >> b;
    if ((lesser >= (b - a + 1) / 2) && ((b - a + 1) % 2 == 0)) {
      cout << 1 << '\n';
    } else {
      cout << 0 << '\n';
    }
    m -= 1;
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

Eugeny has array a = a1, a2, ..., an, consisting of n integers. Each integer ai equals to -1, or to 1. Also, he has m queries:

  * Query number i is given as a pair of integers li, ri (1 ≤ li ≤ ri ≤ n). 
  * The response to the query will be integer 1, if the elements of array a can be rearranged so as the sum ali + ali + 1 + ... + ari = 0, otherwise the response to the query will be integer 0. 



Help Eugeny, answer all his queries.

Input

The first line contains integers n and m (1 ≤ n, m ≤ 2·105). The second line contains n integers a1, a2, ..., an (ai = -1, 1). Next m lines contain Eugene's queries. The i-th line contains integers li, ri (1 ≤ li ≤ ri ≤ n).

Output

Print m integers — the responses to Eugene's queries in the order they occur in the input.

Examples

Input

2 3
1 -1
1 1
1 2
2 2


Output

0
1
0


Input

5 5
-1 1 1 1 -1
1 1
2 3
3 5
2 5
1 5


Output

0
1
0
1
0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int arr[2 * 100500];
int p = 0, ne = 0;
int n, m;
int lesser = min(ne, p);
int a, b;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  cin >> m;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == -1) {
      p += 1;
    }
  }
  ne = abs(p - n);
  lesser = min(ne, p);
  while (m > 0) {
    cin >> a;
    cin >> b;
    if ((lesser >= (b - a + 1) / 2) && ((b - a + 1) % 2 == 0)) {
      cout << 1 << '\n';
    } else {
      cout << 0 << '\n';
    }
    m -= 1;
  }
  return 0;
}

```