#include <bits/stdc++.h>
using namespace std;
int n, m, tag, querries, sum_, pair_;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  for (cin >> n >> querries, m = n; m--;)
    cin >> tag, tag > 0 ? sum_++ : pair_++;
  pair_ = min(pair_, sum_);
  while (querries--) {
    cin >> tag >> m, sum_ = m - tag + 1;
    if (sum_ % 2 == 0 && sum_ / 2 <= pair_)
      cout << 1 << "\n";
    else
      cout << 0 << "\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;
int n, m, tag, querries, sum_, pair_;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  for (cin >> n >> querries, m = n; m--;)
    cin >> tag, tag > 0 ? sum_++ : pair_++;
  pair_ = min(pair_, sum_);
  while (querries--) {
    cin >> tag >> m, sum_ = m - tag + 1;
    if (sum_ % 2 == 0 && sum_ / 2 <= pair_)
      cout << 1 << "\n";
    else
      cout << 0 << "\n";
  }
}

```