#include <bits/stdc++.h>
using namespace std;
const int dx[] = {0, 0, 1, -1, 1, 1, -1, -1};
const int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};
template <typename T>
bool inline between(const T& lbound, const T& value, const T& ubound) {
  return lbound <= value && value <= ubound;
}
int main() {
  int n, m, i, l, r, neg, pos, k, nMax;
  while (scanf("%d %d", &n, &m) != EOF) {
    neg = pos = 0;
    for ((i) = 0; (i) < (n); (i)++) {
      scanf("%d", &k);
      if (k > 0)
        pos++;
      else
        neg++;
    }
    nMax = min(pos, neg);
    for ((i) = 0; (i) < (m); (i)++) {
      scanf("%d %d", &l, &r);
      l--, r--;
      k = (r - l) + 1;
      if (k % 2 == 0 && k / 2 <= nMax)
        printf("1\n");
      else
        printf("0\n");
    }
  }
  fclose(stdout);
  return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
const int dx[] = {0, 0, 1, -1, 1, 1, -1, -1};
const int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};
template <typename T>
bool inline between(const T& lbound, const T& value, const T& ubound) {
  return lbound <= value && value <= ubound;
}
int main() {
  int n, m, i, l, r, neg, pos, k, nMax;
  while (scanf("%d %d", &n, &m) != EOF) {
    neg = pos = 0;
    for ((i) = 0; (i) < (n); (i)++) {
      scanf("%d", &k);
      if (k > 0)
        pos++;
      else
        neg++;
    }
    nMax = min(pos, neg);
    for ((i) = 0; (i) < (m); (i)++) {
      scanf("%d %d", &l, &r);
      l--, r--;
      k = (r - l) + 1;
      if (k % 2 == 0 && k / 2 <= nMax)
        printf("1\n");
      else
        printf("0\n");
    }
  }
  fclose(stdout);
  return 0;
}

```