#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n, m;
  cin >> n >> m;
  int arr[n];
  int a = 0, b = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 1)
      a++;
    else
      b++;
  }
  vector<pair<int, int>> v(m);
  for (int i = 0; i < m; i++) cin >> v[i].first >> v[i].second;
  for (int i = 0; i < m; i++) {
    if ((v[i].second - v[i].first + 1) % 2 == 1)
      cout << "0" << endl;
    else {
      int c = v[i].second - v[i].first + 1;
      int d = c / 2;
      if (d <= a && d <= b)
        cout << "1" << endl;
      else
        cout << "0" << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
void solve() {
  int n, m;
  cin >> n >> m;
  int arr[n];
  int a = 0, b = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 1)
      a++;
    else
      b++;
  }
  vector<pair<int, int>> v(m);
  for (int i = 0; i < m; i++) cin >> v[i].first >> v[i].second;
  for (int i = 0; i < m; i++) {
    if ((v[i].second - v[i].first + 1) % 2 == 1)
      cout << "0" << endl;
    else {
      int c = v[i].second - v[i].first + 1;
      int d = c / 2;
      if (d <= a && d <= b)
        cout << "1" << endl;
      else
        cout << "0" << endl;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

```