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 int MAXN = 100007;
const int MOD = 1000000007;
int ar[MAXN];
int dp[MAXN];
vector<set<int>> vs;
vector<int> idx;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  vs.emplace_back();
  vs.back().insert(0);
  idx.push_back(0);
  int tm = 1;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
    if (vs.back().find(ar[i]) != vs.back().end()) {
      tm += tm;
      if (tm >= MOD) tm -= MOD;
    } else {
      vs.emplace_back();
      for (int x : vs[vs.size() - 2]) {
        vs.back().insert(x);
        vs.back().insert(x ^ ar[i]);
      }
      idx.push_back(i);
    }
    dp[i] = tm;
  }
  while (q--) {
    int l, x;
    cin >> l >> x;
    int i = upper_bound(idx.begin(), idx.end(), l) - idx.begin() - 1;
    if (vs[i].find(x) == vs[i].end())
      cout << 0 << "\n";
    else
      cout << dp[l] << "\n";
  }
  return 0;
}
