#include <bits/stdc++.h>
using namespace std;
const int maxx = 2e5 + 19;
const int mod = 1000000007LL;
const long long inf = 1e9 + 7;
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  int a[maxx] = {};
  int myek = 0, yek = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == -1)
      myek++;
    else if (a[i] == 1)
      yek++;
  }
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    if ((r - l + 1) % 2)
      cout << 0;
    else {
      if (myek >= ((r - l + 1) / 2) && yek >= ((r - l + 1) / 2))
        cout << 1;
      else
        cout << 0;
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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;
const int maxx = 2e5 + 19;
const int mod = 1000000007LL;
const long long inf = 1e9 + 7;
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  int a[maxx] = {};
  int myek = 0, yek = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == -1)
      myek++;
    else if (a[i] == 1)
      yek++;
  }
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    if ((r - l + 1) % 2)
      cout << 0;
    else {
      if (myek >= ((r - l + 1) / 2) && yek >= ((r - l + 1) / 2))
        cout << 1;
      else
        cout << 0;
    }
    cout << endl;
  }
  return 0;
}

```