#include <bits/stdc++.h>
int intcmp(const void *f, const void *s) { return (*(int *)f - *(int *)s); }
int gcd(int a, int b) { return ((b == 0) ? a : gcd(b, a % b)); }
int preprocess() { return 0; }
int main() {
  int cases;
  int i, j, n;
  int a, m;
  preprocess();
  scanf("%d", &n);
  scanf("%d", &m);
  int ones = 0, minus = 0;
  for (i = 0; i < n; ++i) {
    scanf("%d", &a);
    if (a == 1)
      ones++;
    else
      minus++;
  }
  while (m--) {
    int l, r, cnt;
    scanf("%d%d", &l, &r);
    cnt = r - l + 1;
    if (cnt & 1) {
      printf("0\n");
      continue;
    }
    int left = cnt / 2;
    if ((left <= ones) && (left <= minus)) {
      printf("1\n");
      continue;
    } else {
      printf("0\n");
      continue;
    }
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 intcmp(const void *f, const void *s) { return (*(int *)f - *(int *)s); }
int gcd(int a, int b) { return ((b == 0) ? a : gcd(b, a % b)); }
int preprocess() { return 0; }
int main() {
  int cases;
  int i, j, n;
  int a, m;
  preprocess();
  scanf("%d", &n);
  scanf("%d", &m);
  int ones = 0, minus = 0;
  for (i = 0; i < n; ++i) {
    scanf("%d", &a);
    if (a == 1)
      ones++;
    else
      minus++;
  }
  while (m--) {
    int l, r, cnt;
    scanf("%d%d", &l, &r);
    cnt = r - l + 1;
    if (cnt & 1) {
      printf("0\n");
      continue;
    }
    int left = cnt / 2;
    if ((left <= ones) && (left <= minus)) {
      printf("1\n");
      continue;
    } else {
      printf("0\n");
      continue;
    }
  }
  return 0;
}

```