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;
inline int qpow(int b, int e, int m = 998244353) {
  int a = 1;
  for (; e; e >>= 1, b = (long long)b * b % m)
    if (e & 1) a = (long long)a * b % m;
  return a;
}
int n, m, q, k;
int a[300005], b[300005], c[300005];
const int pp[11] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31};
long long MOD = 1000000007;
void solve() {
  long long n, q;
  cin >> n >> q;
  vector<long long> v(n + 1);
  for (long long i = 1; i <= n; i++) cin >> v[i];
  vector<vector<pair<long long, long long>>> queries(n + 1);
  for (long long i = 1; i <= q; i++) {
    long long l, x;
    cin >> l >> x;
    queries[l].push_back(make_pair(x, i));
  }
  vector<long long> ans(q + 1);
  unordered_set<long long> s;
  s.insert(0);
  long long curans = 1;
  for (long long i = 1; i <= n; i++) {
    if (s.find(v[i]) == s.end()) {
      vector<long long> tm(s.size());
      for (auto itr = s.begin(); itr != s.end(); itr++) {
        tm.push_back(*itr);
      }
      for (long long j = 0; j < tm.size(); j++) {
        s.insert(tm[j] ^ v[i]);
      }
    } else {
      curans = (2 * curans) % MOD;
    }
    for (long long j = 0; j < queries[i].size(); j++) {
      if (s.find(queries[i][j].first) == s.end()) {
        ans[queries[i][j].second] = 0;
      } else {
        ans[queries[i][j].second] = curans;
      }
    }
  }
  for (long long i = 1; i <= q; i++) cout << ans[i] << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  t = 1;
  while (t--) solve();
  return 0;
}
