#include <bits/stdc++.h>
int main() {
  int no_of_elements, no_of_queries, no_of_1 = 0, no_of_minus_1 = 0;
  scanf("%d %d", &no_of_elements, &no_of_queries);
  for (int i = 1; i <= no_of_elements; i++) {
    int element_i;
    scanf("%d", &element_i);
    no_of_1 += (element_i == 1);
    no_of_minus_1 += (element_i == -1);
  }
  for (int i = 1; i <= no_of_queries; i++) {
    int left, right, range;
    scanf("%d %d", &left, &right);
    range = right - (left - 1);
    printf(range % 2 == 0 && no_of_1 >= range / 2 && no_of_minus_1 >= range / 2
               ? "1\n"
               : "0\n");
  }
  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>
int main() {
  int no_of_elements, no_of_queries, no_of_1 = 0, no_of_minus_1 = 0;
  scanf("%d %d", &no_of_elements, &no_of_queries);
  for (int i = 1; i <= no_of_elements; i++) {
    int element_i;
    scanf("%d", &element_i);
    no_of_1 += (element_i == 1);
    no_of_minus_1 += (element_i == -1);
  }
  for (int i = 1; i <= no_of_queries; i++) {
    int left, right, range;
    scanf("%d %d", &left, &right);
    range = right - (left - 1);
    printf(range % 2 == 0 && no_of_1 >= range / 2 && no_of_minus_1 >= range / 2
               ? "1\n"
               : "0\n");
  }
  return 0;
}

```