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 l, x, op;
  bool operator<(const node o) const { return l < o.l; }
} q[100005];
int p[100005], a[100005], ans[100005], n, m, cnt;
void ins(int x) {
  for (int i = 20; i >= 0; i--)
    if ((x >> i) & 1) {
      if (!p[i]) {
        p[i] = x, cnt++;
        return;
      } else
        x ^= p[i];
    }
}
bool check(int x) {
  for (int i = 20; i >= 0; i--)
    if ((x >> i) & 1) {
      if (!p[i]) return false;
      x ^= p[i];
    }
  return true;
}
int Pow(int x, int y) {
  int now = 1;
  for (; y; y >>= 1, x = 1LL * x * x % 1000000007)
    if (y & 1) now = 1LL * now * x % 1000000007;
  return now;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++) scanf("%d%d", &q[i].l, &q[i].x), q[i].op = i;
  sort(q + 1, q + m + 1);
  int j = 1;
  for (int i = 1; i <= n; i++) {
    ins(a[i]);
    while (j <= m && q[j].l == i) {
      if (check(q[j].x))
        ans[q[j].op] = Pow(2, i - cnt);
      else
        ans[q[j].op] = 0;
      j++;
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
