#include <bits/stdc++.h>
using namespace std;
int N, M;
int D, B;
int ans[500005];
int main() {
  scanf("%d%d", &N, &M);
  int i, p, l;
  for (i = 0; i < N; ++i) {
    scanf("%d", &l);
    if (l == 1)
      ++D;
    else
      ++B;
  }
  for (i = 0; i < M; ++i) {
    scanf("%d%d", &p, &l);
    if ((l - p + 1) % 2 == 0 && D >= (l - p + 1) / 2 && B >= (l - p + 1) / 2)
      ans[i] = 1;
    else
      ans[i] = 0;
  }
  for (i = 0; i < M; ++i) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Create a solution in cpp for 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>
using namespace std;
int N, M;
int D, B;
int ans[500005];
int main() {
  scanf("%d%d", &N, &M);
  int i, p, l;
  for (i = 0; i < N; ++i) {
    scanf("%d", &l);
    if (l == 1)
      ++D;
    else
      ++B;
  }
  for (i = 0; i < M; ++i) {
    scanf("%d%d", &p, &l);
    if ((l - p + 1) % 2 == 0 && D >= (l - p + 1) / 2 && B >= (l - p + 1) / 2)
      ans[i] = 1;
    else
      ans[i] = 0;
  }
  for (i = 0; i < M; ++i) printf("%d\n", ans[i]);
  return 0;
}

```