#include <bits/stdc++.h>
int main() {
  int i, neg = 0, pos = 0, n, m, x, l, r, h;
  scanf("%d %d", &n, &m);
  for (x = 0; x < n; x++) {
    scanf("%d", &i);
    if (i == -1) {
      neg = neg + 1;
    } else {
      pos = pos + 1;
    }
  }
  for (x = 1; x <= m; x++) {
    scanf("%d %d", &l, &r);
    h = (r - l + 1);
    if (h % 2 == 0 && ((h / 2) <= neg && (h / 2) <= pos)) {
      printf("1\n");
    } else {
      printf("0\n");
    }
  }
}

### Prompt

Please formulate a CPP solution to the following problem:

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>
int main() {
  int i, neg = 0, pos = 0, n, m, x, l, r, h;
  scanf("%d %d", &n, &m);
  for (x = 0; x < n; x++) {
    scanf("%d", &i);
    if (i == -1) {
      neg = neg + 1;
    } else {
      pos = pos + 1;
    }
  }
  for (x = 1; x <= m; x++) {
    scanf("%d %d", &l, &r);
    h = (r - l + 1);
    if (h % 2 == 0 && ((h / 2) <= neg && (h / 2) <= pos)) {
      printf("1\n");
    } else {
      printf("0\n");
    }
  }
}

```