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 long long maxn = 100010, mod = 1e9 + 7;
long long n, q, a[maxn], b[30], last;
long long ans[maxn];
struct node {
  long long l, x, id;
} e[maxn];
bool cmp(node a, node b) { return a.l < b.l; }
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long solve(long long l, long long x) {
  for (long long i = last + 1; i <= l; i++)
    for (long long j = 20; j >= 0; j--)
      if (a[i] >> j & 1) {
        if (!b[j]) {
          b[j] = a[i];
          break;
        } else
          a[i] ^= b[j];
      }
  last = l;
  long long cnt = 0;
  for (long long i = 20; i >= 0; i--) {
    if (b[i]) {
      cnt++;
      if (x >> i & 1) x ^= b[i];
    }
  }
  if (x) return 0;
  return qpow(2, l - cnt);
}
int main() {
  scanf("%lld%lld", &n, &q);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i <= q; i++) {
    scanf("%lld%lld", &e[i].l, &e[i].x);
    e[i].id = i;
  }
  sort(e + 1, e + 1 + q, cmp);
  for (long long i = 1; i <= q; i++) ans[e[i].id] = solve(e[i].l, e[i].x);
  for (long long i = 1; i <= q; i++) printf("%lld\n", ans[i]);
  return 0;
}
