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 chk[(1 << 20) + 5];
int a[(int)(1e5 + 5)], prn[(int)(1e5 + 5)];
pair<int, pair<int, int> > qr[(int)(1e5 + 5)];
vector<int> v;
int main() {
  int n, q, i, j, k, ans, l, s;
  scanf("%d %d", &n, &q);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 0; i < q; i++) {
    scanf("%d %d", &qr[i].first, &qr[i].second.first);
    qr[i].second.second = i;
  }
  sort(qr, qr + q);
  ans = 1;
  chk[0] = 1;
  l = 1;
  v.push_back(0);
  for (i = 0; i < q; i++) {
    j = qr[i].first;
    while (l <= j) {
      if (!chk[a[l]]) {
        s = (int)v.size();
        for (k = 0; k < s; k++) {
          v.push_back(v[k] ^ a[l]);
          chk[v[k] ^ a[l]] = 1;
        }
      } else
        ans = (ans + ans) % (int)(1e9 + 7);
      l++;
    }
    if (chk[qr[i].second.first])
      prn[qr[i].second.second] = ans;
    else
      prn[qr[i].second.second] = 0;
  }
  for (i = 0; i < q; i++) printf("%d\n", prn[i]);
  return 0;
}
