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 mod = 1000000007;
int bases[25];
long long ans[100010];
long long power(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
int ld = 0;
int pp = 1;
void add(int b) {
  for (int i = 0; i < 20; i++)
    if (b >> i & 1) b ^= bases[i];
  if (!b) {
    ld++;
  } else {
    int row = 20;
    while (!(b >> row)) row--;
    for (int i = 0; i < 20; i++)
      if (bases[i] >> row & 1) bases[i] ^= b;
    bases[row] = b;
  }
}
long long count(int b) {
  for (int i = 0; i < 20; i++)
    if (b >> i & 1) b ^= bases[i];
  if (b)
    return 0;
  else
    return power(2LL, (long long)ld);
}
int v[100010];
struct query {
  int i, l, x;
} que[100010];
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  for (int i = 0; i < q; i++) {
    int l, x;
    scanf("%d %d", &l, &x);
    que[i] = {i, l, x};
  }
  sort(que, que + q, [](query a, query b) { return a.l < b.l; });
  int pos = 0;
  for (int i = 0; i < q; i++) {
    while (pos < que[i].l) add(v[pos++]);
    ans[que[i].i] = count(que[i].x);
  }
  for (int i = 0; i < q; i++) printf("%lld\n", ans[i]);
  return 0;
}
