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>
const double Pi = acos(-1.0);
using namespace std;
const long long mod = 1000000007;
const int maxN = 100005;
int n, q;
int a[maxN];
vector<pair<int, int> > qi[maxN];
vector<int> s;
bool in[(1 << 20) + 5];
long long ans[maxN];
int main(int argc, char** argv) {
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int l, x;
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &l, &x);
    qi[l - 1].push_back(pair<int, int>(i, x));
  }
  s.push_back(0);
  in[0] = true;
  long long cur = 1;
  for (int i = 0; i < n; i++) {
    if (!in[a[i]]) {
      int sz = s.size();
      for (int j = 0; j < sz; j++) {
        int tmp = s[j] ^ a[i];
        s.push_back(tmp);
        in[tmp] = true;
      }
    } else
      cur = (cur * 2) % mod;
    for (int j = 0; j < (int)qi[i].size(); j++) {
      if (in[qi[i][j].second]) ans[qi[i][j].first] = cur;
    }
  }
  for (int i = 0; i < q; i++) printf("%lld\n", ans[i]);
  return 0;
}
