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 mod = 1e9 + 7;
int dp[21][100005];
int sum[100005];
int main() {
  memset(dp, -1, sizeof(dp));
  int n, q, a;
  scanf("%d%d", &n, &q);
  sum[0] = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    for (int j = 0; j < 20; j++) dp[j][i] = dp[j][i - 1];
    sum[i] = sum[i - 1];
    bool ok = 0;
    for (int j = 19; j >= 0; j--) {
      if ((a >> j) & 1) {
        if (dp[j][i] == -1) {
          dp[j][i] = a;
          ok = 1;
          break;
        } else
          a ^= dp[j][i];
      }
    }
    if (!ok) sum[i] = (sum[i] * 2) % mod;
  }
  for (int i = 1; i <= q; i++) {
    int l, x;
    scanf("%d%d", &l, &x);
    bool ok = 1;
    for (int j = 19; j >= 0; j--) {
      if ((x >> j) & 1) {
        if (dp[j][l] == -1) {
          ok = 0;
          break;
        } else
          x ^= dp[j][l];
      }
    }
    if (ok)
      printf("%d\n", sum[l]);
    else
      printf("0\n");
  }
}
