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 long long maxn = 1e5 + 5;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 500;
const long long maxm = 21;
pair<long long, pair<long long, long long> > qu[maxn];
long long a[maxn];
long long ans[maxn];
bool vis[1 << maxm][maxm];
long long cnt = 0;
bool find(long long a) { return vis[a][cnt]; }
void add(long long a) {
  for (long long i = 0; i < (1 << maxm); i++) {
    vis[i][cnt + 1] = (vis[i][cnt] | vis[a ^ i][cnt]);
  }
  cnt++;
}
int main() {
  vis[0][0] = 1;
  long long n, q;
  cin >> n >> q;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < q; i++) {
    long long l, x;
    cin >> l >> x;
    l--;
    qu[i] = make_pair(l, make_pair(x, i));
  }
  long long sefr = 1;
  sort(qu, qu + q);
  long long p = 0;
  for (long long i = 0; i < n; i++) {
    if (find(a[i])) {
      sefr = (sefr * 2) % mod;
    } else {
      add(a[i]);
    }
    while (p < q && qu[p].first == i) {
      if (find(qu[p].second.first)) {
        ans[qu[p].second.second] = sefr;
      } else {
        ans[qu[p].second.second] = 0;
      }
      p++;
    }
  }
  for (long long i = 0; i < q; i++) {
    cout << ans[i] << endl;
  }
}
