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 fast_pow(long long x, long long y, long long p);
const long long N = 300005;
long long n, q;
long long ar[N];
vector<vector<pair<long long, long long> > > queries(N);
long long fans[N];
void solve() {
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> ar[i];
  for (long long i = 1; i <= q; i++) {
    long long l, x;
    cin >> l >> x;
    queries[l].push_back({x, i});
  }
  set<long long> s;
  s.insert(0);
  long long ans = 1;
  for (long long i = 1; i <= n; i++) {
    if (s.find(ar[i]) != s.end())
      ans = (ans * 2) % 1000000007;
    else {
      vector<long long> temp;
      for (auto x : s) temp.push_back(x ^ ar[i]);
      for (auto it : temp) s.insert(it);
    }
    for (auto it : queries[i]) {
      if (s.find(it.first) != s.end())
        fans[it.second] = ans;
      else
        fans[it.second] = 0;
    }
  }
  for (long long i = 1; i <= q; i++) cout << fans[i] << "\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  while (test--) {
    solve();
  }
  return 0;
}
long long fast_pow(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
