#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans % m;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m;
  cin >> n >> m;
  long long count1 = 0;
  long long count2 = 0;
  for (long long i = 0; i < n; i++) {
    long long temp;
    cin >> temp;
    if (temp > 0) {
      count2++;
    } else {
      count1++;
    }
  }
  long long pairs = min(count1, count2);
  for (long long i = 0; i < m; i++) {
    long long l, r;
    cin >> l >> r;
    if ((r - l) % 2 != 0) {
      long long check = r - l + 1;
      check = check / 2;
      if (check <= pairs) {
        cout << "1";
      } else {
        cout << "0";
      }
    } else {
      cout << "0";
    }
    cout << "\n";
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans % m;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m;
  cin >> n >> m;
  long long count1 = 0;
  long long count2 = 0;
  for (long long i = 0; i < n; i++) {
    long long temp;
    cin >> temp;
    if (temp > 0) {
      count2++;
    } else {
      count1++;
    }
  }
  long long pairs = min(count1, count2);
  for (long long i = 0; i < m; i++) {
    long long l, r;
    cin >> l >> r;
    if ((r - l) % 2 != 0) {
      long long check = r - l + 1;
      check = check / 2;
      if (check <= pairs) {
        cout << "1";
      } else {
        cout << "0";
      }
    } else {
      cout << "0";
    }
    cout << "\n";
  }
  return 0;
}

```