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 pw[100005], mn[35];
vector<int> bit[35];
int idx[35];
int main() {
  pw[0] = 1;
  for (int i = 1; i <= 100000; i++) pw[i] = (pw[i - 1] * 2) % mod;
  for (int i = 0; i <= 21; i++) mn[i] = 1e9, idx[i] = 1;
  mn[0] = 0;
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    int b = a, cnt = 0;
    while (b) cnt++, b /= 2;
    bit[cnt].push_back(a);
    mn[cnt] = min(mn[cnt], i);
    for (int j = 21; j > 0; j--) {
      while (idx[j] < bit[j].size()) {
        int x = bit[j][0] ^ bit[j][idx[j]];
        int b = x;
        int cnt = 0;
        while (b) cnt++, b /= 2;
        bit[cnt].push_back(x);
        mn[cnt] = min(mn[cnt], i);
        idx[j]++;
      }
    }
  }
  for (int i = 0; i < q; i++) {
    int l, x;
    scanf("%d%d", &l, &x);
    bool ok = 1;
    int cnt = 0;
    for (int j = 20; j >= 0; j--) {
      if (mn[j + 1] <= l) cnt++;
      if ((x >> j) & 1) {
        if (mn[j + 1] <= l) {
          x ^= bit[j + 1][0];
          continue;
        }
        ok = 0;
      }
    }
    int ans = pw[l - cnt];
    if (!ok) ans = 0;
    printf("%d\n", ans);
  }
}
