#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  vector<int> v(n + 1);
  vector<int> neg(n + 1);
  int cnt = 0;
  for (int i(1); i <= n; ++i) {
    cin >> v[i];
    if (v[i] < 0) v[i] = 0, neg[i] = 1;
  }
  int total = accumulate(v.begin(), v.end(), 0);
  int total1 = accumulate(neg.begin(), neg.end(), 0);
  for (int i(2); i <= n; ++i)
    v[i] = v[i] + v[i - 1], neg[i] = neg[i] + neg[i - 1];
  while (m--) {
    int l, r;
    cin >> l >> r;
    int sum = v[r] - v[l - 1];
    int sum1 = neg[r] - neg[l - 1];
    if (sum > sum1) {
      int rem = sum - sum1;
      int baki = total1 - sum1;
      if (rem & 1) {
        cout << 0 << endl;
        continue;
      }
      if (rem / 2 <= baki && (rem / 2))
        cout << 1 << endl;
      else
        cout << 0 << endl;
    } else if (sum == sum1)
      cout << 1 << endl;
    else {
      if (sum < sum1) {
        int rem = sum1 - sum;
        if (rem & 1) {
          cout << 0 << endl;
          continue;
        }
        int baki = total - sum;
        if (rem / 2 <= baki && (rem / 2))
          cout << 1 << endl;
        else
          cout << 0 << endl;
      }
    }
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  vector<int> v(n + 1);
  vector<int> neg(n + 1);
  int cnt = 0;
  for (int i(1); i <= n; ++i) {
    cin >> v[i];
    if (v[i] < 0) v[i] = 0, neg[i] = 1;
  }
  int total = accumulate(v.begin(), v.end(), 0);
  int total1 = accumulate(neg.begin(), neg.end(), 0);
  for (int i(2); i <= n; ++i)
    v[i] = v[i] + v[i - 1], neg[i] = neg[i] + neg[i - 1];
  while (m--) {
    int l, r;
    cin >> l >> r;
    int sum = v[r] - v[l - 1];
    int sum1 = neg[r] - neg[l - 1];
    if (sum > sum1) {
      int rem = sum - sum1;
      int baki = total1 - sum1;
      if (rem & 1) {
        cout << 0 << endl;
        continue;
      }
      if (rem / 2 <= baki && (rem / 2))
        cout << 1 << endl;
      else
        cout << 0 << endl;
    } else if (sum == sum1)
      cout << 1 << endl;
    else {
      if (sum < sum1) {
        int rem = sum1 - sum;
        if (rem & 1) {
          cout << 0 << endl;
          continue;
        }
        int baki = total - sum;
        if (rem / 2 <= baki && (rem / 2))
          cout << 1 << endl;
        else
          cout << 0 << endl;
      }
    }
  }
  return 0;
}

```