Ehab has an array a of n integers. He likes the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) and he likes to bother Mahmoud so he came up with a problem. He gave Mahmoud q queries. In each of them, he gave Mahmoud 2 integers l and x, and asked him to find the number of subsequences of the first l elements of the array such that their bitwise-xor sum is x. Can you help Mahmoud answer the queries?

A subsequence can contain elements that are not neighboring.

Input

The first line contains integers n and q (1 ≤ n, q ≤ 105), the number of elements in the array and the number of queries.

The next line contains n integers a1, a2, ..., an (0 ≤ ai < 220), the elements of the array.

The next q lines, each contains integers l and x (1 ≤ l ≤ n, 0 ≤ x < 220), representing the queries.

Output

For each query, output its answer modulo 109 + 7 in a newline.

Examples

Input

5 5
0 1 2 3 4
4 3
2 0
3 7
5 7
5 8


Output

4
2
0
4
0


Input

3 2
1 1 1
3 1
2 0


Output

4
2

Note

The bitwise-xor sum of the empty set is 0 and the bitwise-xor sum of a set containing one element is that element itself.
#include <bits/stdc++.h>
using namespace std;
const int MAX = 112345;
const int MAX_BASE = 20;
const int MOD = 1e9 + 7;
int data[MAX];
int bases[MAX][MAX_BASE + 1];
void cal(int a, int b[]) {
  for (int j = MAX_BASE; j >= 0; j--) {
    if (a >> j & 1) {
      if (b[j])
        a ^= b[j];
      else {
        b[j] = a;
        for (int k = j - 1; k >= 0; k--)
          if (b[k] && b[j] >> k & 1) b[j] ^= b[k];
        for (int k = j + 1; k <= MAX_BASE; k++)
          if (b[k] >> j & 1) b[k] ^= b[j];
        break;
      }
    }
  }
}
long long pow_num[MAX];
int check(int a, int b[], int l) {
  int rst = 0;
  int tot = 0;
  for (int i = MAX_BASE; i >= 0; i--) {
    if (a >> i & 1) rst ^= b[i];
    if (b[i]) tot++;
  }
  if (rst != a) return 0;
  return pow_num[l - tot];
}
int main() {
  pow_num[0] = 1;
  for (int i = 1; i < MAX; i++) pow_num[i] = pow_num[i - 1] * 2 % MOD;
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> data[i];
  for (int i = 1; i <= n; i++) {
    memcpy(bases[i], bases[i - 1], sizeof(bases[i]));
    cal(data[i], bases[i]);
  }
  while (q--) {
    int l, x;
    cin >> l >> x;
    cout << check(x, bases[l], l) << endl;
  }
}
