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;
struct Node {
  int no, l;
  long long x;
  bool friend operator<(Node x1, Node x2) { return x1.l < x2.l; }
} q[100005];
long long init[100005];
long long num[100005];
long long ans[100005];
long long temp[25];
bool ok(long long x) {
  for (int i = 19; i >= 0; i--) {
    if (x & init[i]) x ^= temp[i];
  }
  return x == 0;
}
int main() {
  init[0] = 1;
  for (int i = 1; i <= 100000; i++) init[i] = (init[i - 1] * 2) % 1000000007;
  int n, m, now = 0, cnt = 0;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld", &num[i]);
  for (int i = 1; i <= m; i++) {
    q[i].no = i;
    scanf("%d%lld", &q[i].l, &q[i].x);
  }
  sort(q + 1, q + 1 + m);
  for (int i = 1; i <= m; i++) {
    while (now < q[i].l) {
      now++;
      for (int j = 19; j >= 0; j--) {
        if (num[now] & init[j]) {
          if (temp[j] == 0) {
            temp[j] = num[now];
            cnt++;
            break;
          } else
            num[now] ^= temp[j];
        }
      }
    }
    if (ok(q[i].x)) ans[q[i].no] = init[now - cnt];
  }
  for (int i = 1; i <= m; i++) printf("%lld\n", ans[i]);
  return 0;
}
