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 int N = 100005;
const int MOD = 1000000007;
int n, q, bin[N], bas[N][25], cnt[N];
bool check(int l, int x) {
  for (int i = 20; i >= 0; i--)
    if (x & bin[i]) {
      if (!bas[l][i]) return 0;
      x ^= bas[l][i];
    }
  return x == 0;
}
int main() {
  scanf("%d%d", &n, &q);
  bin[0] = 1;
  for (int i = 1; i <= max(n, 20); i++) bin[i] = bin[i - 1] * 2 % MOD;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 20; j++) bas[i][j] = bas[i - 1][j];
    cnt[i] = cnt[i - 1];
    int x;
    scanf("%d", &x);
    for (int j = 20; j >= 0; j--)
      if (x & bin[j]) {
        if (!bas[i][j]) {
          bas[i][j] = x;
          cnt[i]++;
          break;
        } else
          x ^= bas[i][j];
      }
  }
  while (q--) {
    int l, x;
    scanf("%d%d", &l, &x);
    if (!check(l, x))
      puts("0");
    else
      printf("%d\n", bin[l - cnt[l]]);
  }
  return 0;
}
