#include <bits/stdc++.h>
using namespace std;
int main(void) {
  int n, t;
  scanf("%d %d", &n, &t);
  int x, i, a, b, cnt1 = 0, cnt2 = 0;
  for (i = 0; i < n; ++i) {
    scanf("%d", &x);
    if (x == 1)
      ++cnt1;
    else
      ++cnt2;
  }
  while (t--) {
    scanf("%d %d", &a, &b);
    int len = (abs(a - b) + 1);
    if (len % 2)
      cout << "0\n";
    else {
      if (cnt1 >= (len / 2) && cnt2 >= (len / 2))
        cout << "1\n";
      else
        cout << "0\n";
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 main(void) {
  int n, t;
  scanf("%d %d", &n, &t);
  int x, i, a, b, cnt1 = 0, cnt2 = 0;
  for (i = 0; i < n; ++i) {
    scanf("%d", &x);
    if (x == 1)
      ++cnt1;
    else
      ++cnt2;
  }
  while (t--) {
    scanf("%d %d", &a, &b);
    int len = (abs(a - b) + 1);
    if (len % 2)
      cout << "0\n";
    else {
      if (cnt1 >= (len / 2) && cnt2 >= (len / 2))
        cout << "1\n";
      else
        cout << "0\n";
    }
  }
  return 0;
}

```