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 arr[200005], ans[200005];
map<long long, long long> memo;
vector<pair<long long, long long> > V[200005];
vector<long long> t;
int main() {
  long long n, q;
  scanf("%lld %lld", &n, &q);
  for (long long i = 1; i <= n; i++) scanf("%lld", &arr[i]);
  for (long long i = 1; i <= q; i++) {
    long long pos, val;
    scanf("%lld %lld", &pos, &val);
    V[pos].push_back(make_pair(i, val));
  }
  t.push_back(0);
  memo[0] = 1;
  long long f = 1;
  for (long long i = 1; i <= n; i++) {
    long long val = arr[i];
    if (memo[val]) {
      f = (f * 2) % 1000000007;
    } else {
      vector<long long> toadd;
      for (long long j = 0; j < t.size(); j++) {
        toadd.push_back(t[j] ^ val);
      }
      for (long long j = 0; j < toadd.size(); j++) {
        t.push_back(toadd[j]);
        memo[toadd[j]] = 1;
      }
    }
    for (long long j = 0; j < V[i].size(); j++) {
      long long pos = V[i][j].first;
      long long val = V[i][j].second;
      if (memo[val])
        ans[pos] = f;
      else
        ans[pos] = 0;
    }
  }
  for (long long i = 1; i <= q; i++) printf("%lld\n", ans[i]);
}
