#include <bits/stdc++.h>
int main() {
  int str[200005] = {0}, n, m;
  int i, k, a, b, n1, n2, t;
  while (scanf("%d%d", &n, &m) != EOF) {
    n1 = n2 = t = k = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &str[i]);
      if (str[i] == 1)
        n1++;
      else if (str[i] == -1)
        n2++;
    }
    if (n1 >= n2)
      t = n2;
    else
      t = n1;
    for (i = 1; i <= m; i++) {
      scanf("%d%d", &a, &b);
      k = b - a + 1;
      if (k <= (2 * t) && (k % 2 == 0))
        printf("1\n");
      else
        printf("0\n");
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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>
int main() {
  int str[200005] = {0}, n, m;
  int i, k, a, b, n1, n2, t;
  while (scanf("%d%d", &n, &m) != EOF) {
    n1 = n2 = t = k = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &str[i]);
      if (str[i] == 1)
        n1++;
      else if (str[i] == -1)
        n2++;
    }
    if (n1 >= n2)
      t = n2;
    else
      t = n1;
    for (i = 1; i <= m; i++) {
      scanf("%d%d", &a, &b);
      k = b - a + 1;
      if (k <= (2 * t) && (k % 2 == 0))
        printf("1\n");
      else
        printf("0\n");
    }
  }
  return 0;
}

```