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;
struct query {
  int x, i;
  query() {}
  query(int x, int i) : x(x), i(i) {}
};
int n, q, arr[100000], ans[100000], has[1 << 20];
vector<query> qa[100000];
vector<int> t;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", arr + i);
  for (int i = 0; i < q; i++) {
    int l, x;
    scanf("%d%d", &l, &x);
    l--;
    qa[l].push_back(query(x, i));
  }
  int val = 1;
  has[0] = 1;
  t.push_back(0);
  for (int i = 0; i < n; i++) {
    if (has[arr[i]])
      val = (val * 2) % mod;
    else {
      int s = t.size();
      for (int j = 0; j < s; j++) {
        has[arr[i] ^ t[j]] = 1;
        t.push_back(arr[i] ^ t[j]);
      }
    }
    for (auto &it : qa[i]) {
      if (has[it.x])
        ans[it.i] = val;
      else
        ans[it.i] = 0;
    }
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}
