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 n, q;
const int maxm = 1e5 + 10;
vector<pair<int, int> > que[maxm];
int num[maxm];
vector<int> has;
bool vis[1 << 20];
int answer[maxm];
int main() {
  scanf("%d%d", &n, &q);
  has.push_back(0);
  vis[0] = true;
  for (int i = 1; i <= n; i++) {
    scanf("%d", num + i);
  }
  for (int i = 0; i < q; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    que[a].push_back(make_pair(i, b));
  }
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    if (vis[num[i]])
      ans = (ans << 1) % mod;
    else {
      int si = has.size();
      for (int j = 0; j < si; j++) {
        has.push_back(has[j] ^ num[i]);
        vis[has.back()] = true;
      }
    }
    for (pair<int, int> now : que[i]) {
      answer[now.first] = (ans * vis[now.second]) % mod;
    }
  }
  for (int i = 0; i < q; i++) {
    printf("%d ", answer[i]);
  }
  return 0;
}
