#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int a[N], n, m, ps[N], l, r, t1[N], t0[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    t1[i] = t1[i - 1] + (a[i] == 1);
    t0[i] = t0[i - 1] + (a[i] == -1);
  }
  while (m--) {
    cin >> l >> r;
    int x, y;
    x = t1[r] - t1[l - 1];
    y = t0[r] - t0[l - 1];
    if (abs(x - y) % 2 == 0) {
      if (x - y >= 0 && (x - y) / 2 <= t0[n] - y)
        cout << 1 << endl;
      else if (x - y <= 0 && abs(x - y) / 2 <= t1[n] - x)
        cout << 1 << endl;
      else
        cout << 0 << endl;
    } else
      cout << 0 << endl;
  }
  return 0;
}

### Prompt

Generate 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 N = 2e5 + 10;
int a[N], n, m, ps[N], l, r, t1[N], t0[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    t1[i] = t1[i - 1] + (a[i] == 1);
    t0[i] = t0[i - 1] + (a[i] == -1);
  }
  while (m--) {
    cin >> l >> r;
    int x, y;
    x = t1[r] - t1[l - 1];
    y = t0[r] - t0[l - 1];
    if (abs(x - y) % 2 == 0) {
      if (x - y >= 0 && (x - y) / 2 <= t0[n] - y)
        cout << 1 << endl;
      else if (x - y <= 0 && abs(x - y) / 2 <= t1[n] - x)
        cout << 1 << endl;
      else
        cout << 0 << endl;
    } else
      cout << 0 << endl;
  }
  return 0;
}

```