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 mod = 1e9 + 7;
int n, q, a[110000];
vector<pair<int, int> > Q[110000];
bool b[1 << 20];
vector<int> jj;
int ans[110000];
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= q; i++) {
    int l, x;
    scanf("%d%d", &l, &x);
    Q[l].push_back(make_pair(x, i));
  }
  int aa = 1;
  b[0] = 1;
  jj.push_back(0);
  for (int i = 1; i <= n; i++) {
    if (b[a[i]]) {
      aa *= 2;
      aa %= mod;
    } else {
      int m = jj.size();
      for (int j = 0; j < m; j++) {
        jj.push_back(a[i] ^ jj[j]);
        b[a[i] ^ jj[j]] = 1;
      }
    }
    for (auto j = Q[i].begin(); j != Q[i].end(); j++) {
      ans[(*j).second] = aa * b[(*j).first];
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
