#include <bits/stdc++.h>
using namespace std;
int code() {
  long long n, m;
  cin >> n >> m;
  int ar[n];
  long long countn = 0, countp = 0;
  for (long long i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x < 0)
      countn++;
    else
      countp++;
  }
  while ((m--)) {
    long long n1, n2;
    cin >> n1 >> n2;
    long long ran = n2 - n1 + 1;
    if (ran % 2)
      cout << "0";
    else {
      if ((countn >= ran / 2) && (countp >= (ran / 2)))
        cout << "1";
      else
        cout << "0";
    }
    cout << "\n";
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) code();
  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;
int code() {
  long long n, m;
  cin >> n >> m;
  int ar[n];
  long long countn = 0, countp = 0;
  for (long long i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x < 0)
      countn++;
    else
      countp++;
  }
  while ((m--)) {
    long long n1, n2;
    cin >> n1 >> n2;
    long long ran = n2 - n1 + 1;
    if (ran % 2)
      cout << "0";
    else {
      if ((countn >= ran / 2) && (countp >= (ran / 2)))
        cout << "1";
      else
        cout << "0";
    }
    cout << "\n";
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) code();
  return 0;
}

```