#include <bits/stdc++.h>
using namespace std;
int a[200001];
int main() {
  int N, M, i;
  scanf("%d%d", &N, &M);
  int sum1 = 0, sum2 = 0;
  for (i = 1; i <= N; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      sum1++;
    else
      sum2++;
  }
  while (M--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int count = r - l + 1;
    if (count % 2 == 1)
      cout << "0\n";
    else {
      int ans = min(sum1, sum2);
      if (2 * ans >= count)
        cout << "1\n";
      else
        cout << "0\n";
    }
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 a[200001];
int main() {
  int N, M, i;
  scanf("%d%d", &N, &M);
  int sum1 = 0, sum2 = 0;
  for (i = 1; i <= N; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      sum1++;
    else
      sum2++;
  }
  while (M--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int count = r - l + 1;
    if (count % 2 == 1)
      cout << "0\n";
    else {
      int ans = min(sum1, sum2);
      if (2 * ans >= count)
        cout << "1\n";
      else
        cout << "0\n";
    }
  }
  return 0;
}

```