#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1000000000")
using namespace std;
const int size = 300010;
const long long alphabet = 130;
const long long INF = 1000000000;
const double pi = 4 * atan(1.0);
const long long MOD = 1000000007;
double a[30][30], Mi, M0, M, s[30];
int main() {
  int od = 0, minusod = 0;
  int n, m, x;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> x;
    if (x == 1)
      ++od;
    else
      ++minusod;
  }
  int l, r;
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &l, &r);
    int dov = r - l;
    if (!(dov & 1))
      printf("0\n");
    else {
      int lol = (r - l + 1) / 2;
      if (lol <= od && lol <= minusod)
        printf("1\n");
      else
        printf("0\n");
    }
  }
  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>
#pragma comment(linker, "/STACK:1000000000")
using namespace std;
const int size = 300010;
const long long alphabet = 130;
const long long INF = 1000000000;
const double pi = 4 * atan(1.0);
const long long MOD = 1000000007;
double a[30][30], Mi, M0, M, s[30];
int main() {
  int od = 0, minusod = 0;
  int n, m, x;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> x;
    if (x == 1)
      ++od;
    else
      ++minusod;
  }
  int l, r;
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &l, &r);
    int dov = r - l;
    if (!(dov & 1))
      printf("0\n");
    else {
      int lol = (r - l + 1) / 2;
      if (lol <= od && lol <= minusod)
        printf("1\n");
      else
        printf("0\n");
    }
  }
  return 0;
}

```