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 rua {
  int l, x, id;
} q[100001];
int n, Q, _, cnt, a[100001], v[21], res[100001];
bool cmp(rua x, rua y) { return x.l < y.l; }
long long qow(long long x, long long y) {
  return y ? (y & 1 ? x * qow(x, y - 1) % 1000000007
                    : qow(x * x % 1000000007, y / 2))
           : 1;
}
void insert(int x) {
  int k = 0, p = -1;
  while ((1 << k) <= x) k++;
  k--;
  for (int i = k; i >= 0; i--)
    if (v[i])
      x = min(x, x ^ v[i]);
    else if ((1 << i) <= x) {
      v[i] = x, p = i;
      break;
    }
  if (p != -1) {
    for (int i = p + 1; i < 20; i++) v[i] = min(v[i], v[i] ^ v[p]);
    for (int i = p - 1; i >= 0; i--) v[p] = min(v[p], v[i] ^ v[p]);
  }
  if (!x || (p != -1 && !v[p])) cnt++;
}
bool find(int x) {
  for (int i = 19; i >= 0; i--) x = min(x, x ^ v[i]);
  return !x;
}
int main() {
  scanf("%d%d", &n, &Q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= Q; i++) scanf("%d%d", &q[i].l, &q[i].x), q[i].id = i;
  sort(q + 1, q + Q + 1, cmp);
  for (int i = 1; i <= Q; i++) {
    while (_ < q[i].l) insert(a[++_]);
    res[q[i].id] = find(q[i].x) ? qow(2, cnt) : 0;
  }
  for (int i = 1; i <= Q; i++) printf("%d\n", res[i]);
  return 0;
}
