#include <bits/stdc++.h>
using namespace std;
int n, m;
int a[200000];
void run(int casenr) {
  scanf("%d%d", &n, &m);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  int cntpos = 0, cntneg = 0;
  for (int i = (0); i < (n); ++i)
    if (a[i] < 0)
      ++cntneg;
    else
      ++cntpos;
  for (int i = (0); i < (m); ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    int cnt = r - l + 1;
    printf("%d\n",
           cnt % 2 == 0 && cntpos >= cnt / 2 && cntneg >= cnt / 2 ? 1 : 0);
  }
}
int main() {
  run(1);
  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>
using namespace std;
int n, m;
int a[200000];
void run(int casenr) {
  scanf("%d%d", &n, &m);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  int cntpos = 0, cntneg = 0;
  for (int i = (0); i < (n); ++i)
    if (a[i] < 0)
      ++cntneg;
    else
      ++cntpos;
  for (int i = (0); i < (m); ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    int cnt = r - l + 1;
    printf("%d\n",
           cnt % 2 == 0 && cntpos >= cnt / 2 && cntneg >= cnt / 2 ? 1 : 0);
  }
}
int main() {
  run(1);
  return 0;
}

```