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;
int a[100001], c[100001], p2[100001];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 0) continue;
    int k;
    for (k = 20 - 1; k > -1 && (a[i] & (1 << k)) == 0; --k)
      ;
    for (int j = i + 1; j <= n; ++j) {
      if ((a[j] & (1 << k)) > 0) a[j] ^= a[i];
    }
    c[i] = k;
  }
  p2[0] = 1;
  for (int i = 1; i < 100001; ++i) p2[i] = (2ll * p2[i - 1]) % 1000000007;
  vector<int> b;
  for (int i = 1; i <= n; ++i) {
    if (a[i]) {
      b.push_back(i);
    }
  }
  while (q--) {
    int l, x, k = 0;
    scanf("%d%d", &l, &x);
    for (auto i : b) {
      if (i > l) break;
      if ((x & (1 << c[i])) > 0) x ^= a[i];
      ++k;
    }
    int ans = (x == 0) ? p2[l - k] : 0;
    printf("%d\n", ans);
  }
  return 0;
}
