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;
using ll = long long;
const int maxn = 1 << 20, mod = 1e9 + 7;
struct dsu {
  vector<int> sz, par;
  dsu(int n) : sz(n, 1), par(n) { iota((par).begin(), (par).end(), 0); }
  int get(int v) { return (par[v] == v ? v : par[v] = get(par[v])); }
  void unite(int u, int v) {
    if ((u = get(u)) != (v = get(v))) {
      if (sz[u] < sz[v]) swap(u, v);
      par[v] = u;
      sz[u] += sz[v];
    }
  }
};
int n, q, a[maxn], ans[maxn];
vector<pair<int, int>> ve[maxn];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < q; i++) {
    int l, x;
    cin >> l >> x;
    ve[l - 1].push_back({x, i});
  }
  dsu u(maxn);
  int pw = 1;
  for (int i = 0; i < n; i++) {
    bool same = (u.get(0) == u.get(a[i]));
    if (same) {
      pw = pw * 2 % mod;
    } else {
      for (int j = 0; j < maxn; j++) u.unite(j, j ^ a[i]);
    }
    for (pair<int, int> x : ve[i])
      if (u.get(0) == u.get(x.first))
        ans[x.second] = pw;
      else
        ans[x.second] = 0;
  }
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
}
