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 {
  long long l;
  long long x;
  long long num;
  bool operator<(const node n1) const { return l < n1.l; }
} node[200005];
long long p[70];
long long ans;
long long qsm(long long x, long long y) {
  long long a = 1;
  long long base = x;
  while (y > 0) {
    if (y & 1) {
      a = (a * base) % 1000000007;
    }
    base = (base * base) % 1000000007;
    y >>= 1;
  }
  return a;
}
void xxj(long long x) {
  for (int i = 62; i >= 0; i--) {
    if (x & (1ll * 1 << i)) {
      if (!p[i]) {
        p[i] = x;
        break;
      } else {
        x ^= p[i];
      }
    }
  }
  if (x == 0) ans++;
  return;
}
long long query(long long x) {
  for (int i = 62; i >= 0; i--) {
    if (x & (1ll * 1 << i)) {
      x ^= p[i];
    }
  }
  if (x == 0) {
    return qsm(2, ans);
  } else {
    return 0;
  }
}
long long n, q;
long long a[200005];
long long res[200005];
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= q; i++) {
    cin >> node[i].l >> node[i].x;
    node[i].num = i;
  }
  sort(node + 1, node + q + 1);
  long long now = 0;
  for (int i = 1; i <= q; i++) {
    while (now < node[i].l) {
      now++;
      xxj(a[now]);
    }
    res[node[i].num] = query(node[i].x);
  }
  for (int i = 1; i <= q; i++) {
    cout << res[i] << endl;
  }
}
