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 base = 361;
const long long mod = 1e9 + 7;
const long double eps = 1e-5;
const long long maxn = 1e6;
long long n, q, a[200009], basis[100009][23], l, r, i, j, sizet[200009];
void update(long long x, long long y) {
  for (long long ii = 20; ii >= 1; ii--) {
    if (basis[x][ii] == 0) {
      if (((y >> (ii - 1)) & 1) == 1) {
        basis[x][ii] = y;
        sizet[x]++;
        return;
      }
    } else if (((y >> (ii - 1)) & 1) == 1) {
      y ^= basis[x][ii];
    }
  }
}
bool check(long long x, long long y) {
  long long ans = 0;
  for (long long ii = 20; ii > 0; ii--) {
    if (basis[x][ii] != 0) {
      if (((y >> (ii - 1)) & 1) != ((ans >> (ii - 1)) & 1)) {
        ans ^= basis[x][ii];
      }
    }
  }
  return ans == y;
}
long long GTN(long long x, long long y) {
  if (y == 0) {
    return 1;
  } else {
    long long tmp = GTN(x, y / 2);
    if (y % 2 == 0) {
      return (tmp * tmp) % mod;
    } else {
      return (((tmp * tmp) % mod) * x) % mod;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (i = 1; i <= n; i++) {
    sizet[i] = sizet[i - 1];
    for (j = 1; j <= 20; j++) {
      basis[i][j] = basis[i - 1][j];
    }
    cin >> a[i];
    update(i, a[i]);
  }
  for (i = 1; i <= q; i++) {
    cin >> l >> r;
    if (!check(l, r)) {
      cout << 0 << '\n';
    } else {
      cout << GTN(2, l - sizet[l]) << '\n';
    }
  }
}
