#include <bits/stdc++.h>
void f(int ne, int po, int len);
int main() {
  int n, m, i, ne, po, len;
  ne = po = 0;
  scanf("%d%d", &n, &m);
  int a[100], b[200000][2];
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      po++;
    else
      ne++;
  }
  for (i = 0; i < m; i++) scanf("%d%d", &b[i][0], &b[i][1]);
  for (i = 0; i < m; i++) {
    len = b[i][1] - b[i][0] + 1;
    if (!i)
      f(ne, po, len);
    else {
      printf("\n");
      f(ne, po, len);
    }
  }
  return 0;
}
void f(int ne, int po, int len) {
  if (len % 2)
    printf("0");
  else {
    int min;
    min = ne > po ? po : ne;
    if (len / 2 <= min)
      printf("1");
    else
      printf("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>
void f(int ne, int po, int len);
int main() {
  int n, m, i, ne, po, len;
  ne = po = 0;
  scanf("%d%d", &n, &m);
  int a[100], b[200000][2];
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      po++;
    else
      ne++;
  }
  for (i = 0; i < m; i++) scanf("%d%d", &b[i][0], &b[i][1]);
  for (i = 0; i < m; i++) {
    len = b[i][1] - b[i][0] + 1;
    if (!i)
      f(ne, po, len);
    else {
      printf("\n");
      f(ne, po, len);
    }
  }
  return 0;
}
void f(int ne, int po, int len) {
  if (len % 2)
    printf("0");
  else {
    int min;
    min = ne > po ? po : ne;
    if (len / 2 <= min)
      printf("1");
    else
      printf("0");
  }
}

```