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 mo = 1000000007;
const int N = 200005;
int power(int x, int y) {
  int s = 1;
  for (; y; y /= 2, x = 1ll * x * x % mo)
    if (y & 1) s = 1ll * s * x % mo;
  return s;
}
int n, Q, x, y;
int a[N], b[25], ans[N];
vector<pair<int, int> > vec[N];
void insert(int v) {
  for (int i = (int)(20); i >= (int)(0); i--) {
    if ((v & (1 << i)) && !b[i]) b[i] = v;
    if ((v ^ b[i]) < v) v ^= b[i];
  }
}
int main() {
  scanf("%d%d", &n, &Q);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i]);
  for (int i = (int)(1); i <= (int)(Q); i++)
    scanf("%d%d", &x, &y), vec[x].push_back(pair<int, int>(y, i));
  for (int i = (int)(1); i <= (int)(n); i++) {
    insert(a[i]);
    for (auto j : vec[i]) {
      int v = j.first, S = i;
      for (int p = (int)(20); p >= (int)(0); p--) {
        if (b[p]) --S;
        if ((v ^ b[p]) < v) v ^= b[p];
      }
      if (!v) ans[j.second] = power(2, S);
    }
  }
  for (int i = (int)(1); i <= (int)(Q); i++) printf("%d\n", ans[i]);
}
