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;
int INF = 1000000005;
long long INFF = 1000000000000000005;
double PI = acos(-1);
void start() {}
long long n, q;
long long ar[300005];
vector<vector<pair<long long, long long>>> queries(300005);
long long fans[300005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  start();
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> ar[i];
  }
  for (int 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 (int 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 (int i = 1; i <= q; ++i) {
    cout << fans[i] << endl;
  }
  return 0;
}
