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 N = 100005, P = 1000000007;
int n, q, cnt, c[25], a[N], pw[N], ans[N];
void ins(int num) {
  for (int i = 19; i >= 0; --i)
    if (num & (1 << i)) {
      if (!c[i]) {
        c[i] = num;
        ++cnt;
        return;
      } else
        num ^= c[i];
    }
}
bool check(int num) {
  for (int i = 19; i >= 0; --i)
    if (num & (1 << i)) num ^= c[i];
  return !num;
}
struct Query {
  int l, x, id;
} qq[N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q;
  pw[0] = 1;
  for (int i = 1; i <= n; i++) cin >> a[i], pw[i] = pw[i - 1] * 2 % P;
  for (int i = 1; i <= q; i++) cin >> qq[i].l >> qq[i].x, qq[i].id = i;
  sort(qq + 1, qq + q + 1, [](Query a, Query b) { return a.l < b.l; });
  int ptr = 1;
  for (int i = 1; i <= q; i++) {
    while (ptr <= qq[i].l) ins(a[ptr++]);
    if (check(qq[i].x)) ans[qq[i].id] = pw[ptr - 1 - cnt];
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << endl;
  return 0;
}
