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;
long long int d[100005];
long long int a[100005];
const long long int K = 1e9 + 7;
long long int n, q;
vector<long long int> v;
bool vs[(1LL << 21) + 10];
struct query {
  long long int l, x, stt;
  query() { l = x = stt = 0; };
  query(long long int a, long long int b, long long int c) {
    l = a;
    x = b;
    stt = c;
  }
};
bool operator<(query a, query b) { return a.l < b.l; }
vector<query> Q;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= q; i++) {
    long long int l, x;
    cin >> l >> x;
    Q.push_back(query(l, x, i));
  }
  sort(Q.begin(), Q.end());
  long long int res = 1;
  v.push_back(0);
  vs[0] = 1;
  long long int X = 0;
  for (int i = 1; i <= n; i++) {
    if (vs[a[i]]) {
      res = res * 2 % K;
    } else {
      vector<long long int> v1;
      for (auto j : v) {
        v1.push_back(j ^ a[i]);
        vs[j ^ a[i]] = 1;
      }
      for (auto j : v1) v.push_back(j);
    }
    while (Q[X].l == i) {
      if (vs[Q[X].x])
        d[Q[X].stt] = res;
      else
        d[Q[X].stt] = 0;
      X++;
    }
  }
  for (int i = 1; i <= q; i++) cout << d[i] << endl;
  return 0;
}
