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 n, q, x[100010], in[1 << 20], ans[100010], val = 1, l, y;
vector<int> inside;
vector<pair<int, int> > queries[100010];
int main() {
  scanf("%d %d", &n, &q);
  in[0] = 1;
  inside.push_back(0);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x[i]);
  }
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &l, &y);
    queries[l - 1].emplace_back(y, i);
  }
  for (int i = 0; i < n; i++) {
    if (in[x[i]]) {
      val *= 2;
      if (val >= 1000000007) val -= 1000000007;
    } else {
      for (int j = inside.size() - 1; j >= 0; j--) {
        assert(!in[x[i] ^ inside[j]]);
        in[x[i] ^ inside[j]] = 1;
        inside.push_back(x[i] ^ inside[j]);
      }
    }
    for (auto qry : queries[i]) {
      if (in[qry.first])
        ans[qry.second] = val;
      else
        ans[qry.second] = 0;
    }
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}
