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 P[1000000];
int cur;
long long ret[1000000];
int a[1000000];
vector<pair<int, int> > vec[1000000];
struct Base {
  int d[22 + 1];
  int p[22 + 1];
  int sz;
  int cnt;
  Base() {
    sz = 0;
    cnt = 0;
    for (int i = 0; i <= 22; i++) d[i] = p[i] = 0;
  }
  bool insert(int x) {
    int i;
    for (i = 22; ~i; i--) {
      if (!(x & (1LL << i))) continue;
      if (!d[i]) {
        cnt++;
        d[i] = x;
        break;
      }
      x ^= d[i];
    }
    if (x) p[sz++] = x;
    return x > 0;
  }
  int query_max() {
    int i;
    int res = 0;
    for (i = 22; ~i; i--) {
      if ((res ^ d[i]) > res) res ^= d[i];
    }
    return res;
  }
  int query_min() {
    for (int i = 0; i <= 22; i++) {
      if (d[i]) return d[i];
    }
    return 0;
  }
  int getX(int X) {
    for (int i = 22; ~i; i--) {
      if ((1 << i) & X) X ^= d[i];
    }
    if (X == 0)
      return P[cur - sz];
    else
      return 0;
  }
};
int main() {
  int n, q;
  cin >> n >> q;
  P[0] = 1;
  for (int i = 1; i <= n; i++) P[i] = (P[i - 1] * 2) % 1000000007;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < q; i++) {
    int l, x;
    cin >> l >> x;
    vec[l].push_back({x, i});
  }
  Base b;
  for (int i = 1; i <= n; i++) {
    cur = i;
    b.insert(a[i]);
    for (auto p : vec[i]) {
      ret[p.second] = b.getX(p.first);
    }
  }
  for (int i = 0; i < q; i++) cout << ret[i] << endl;
}
