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;
long long i, j, k, l, n, m;
vector<long long> a[200009];
long long mpow(long long n, long long m) {
  long long res = 1, curr = n;
  while (m) {
    if (m & 1) res = (res * curr) % 1000000007;
    curr = (curr * curr) % 1000000007;
    m /= 2;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    a[i] = a[i - 1];
    cin >> k;
    for (auto to : a[i]) {
      k = min(k, k ^ to);
    }
    if (k) a[i].push_back(k);
  }
  while (m--) {
    cin >> k >> l;
    for (auto to : a[k]) l = min(l, l ^ to);
    if (l != 0)
      cout << 0 << '\n';
    else
      cout << mpow(2, k - a[k].size()) << '\n';
  }
  exit(0);
}
