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;
bool ta[2000006];
int main() {
  int n, q;
  int a[100005];
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  vector<pair<pair<int, int>, int>> qrys;
  for (int i = 0; i < q; i++) {
    int l, x;
    scanf("%d %d", &l, &x);
    qrys.push_back({{l, x}, i});
  }
  sort(qrys.begin(), qrys.end());
  int id = 0;
  long long ans[100005];
  vector<int> tao = {0};
  ta[0] = 1;
  long long qtd = 1;
  for (int i = 0; i < n; i++) {
    if (!ta[a[i]]) {
      vector<int> add;
      for (int j : tao) {
        if (!ta[a[i] ^ j]) add.push_back(a[i] ^ j);
        ta[a[i] ^ j] = 1;
      }
      for (int j : add) tao.push_back(j);
    } else
      qtd = (qtd * 2) % 1000000007;
    while (id < qrys.size() && qrys[id].first.first - 1 == i) {
      ans[qrys[id].second] = ta[qrys[id].first.second] ? qtd : 0;
      id++;
    }
  }
  for (int i = 0; i < q; i++) printf("%lld\n", ans[i]);
  return 0;
}
