#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long n, q;
  cin >> n >> q;
  vector<long long> a(n);
  vector<long long> cnt(2);
  for (long long& e : a) {
    cin >> e;
    if (e == -1)
      cnt[0]++;
    else
      cnt[1]++;
  }
  while (q--) {
    long long l, r;
    cin >> l >> r;
    long long sz = (r - l + 1);
    if (sz % 2)
      cout << 0;
    else {
      if (cnt[0] >= sz / 2 && cnt[1] >= sz / 2)
        cout << 1;
      else
        cout << 0;
    }
    cout << "\n";
  }
}

### 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 long long N = 1e5 + 9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long n, q;
  cin >> n >> q;
  vector<long long> a(n);
  vector<long long> cnt(2);
  for (long long& e : a) {
    cin >> e;
    if (e == -1)
      cnt[0]++;
    else
      cnt[1]++;
  }
  while (q--) {
    long long l, r;
    cin >> l >> r;
    long long sz = (r - l + 1);
    if (sz % 2)
      cout << 0;
    else {
      if (cnt[0] >= sz / 2 && cnt[1] >= sz / 2)
        cout << 1;
      else
        cout << 0;
    }
    cout << "\n";
  }
}

```