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;
vector<long long> val(100000 + 5);
struct Gaussian {
  long long no_of_bits = 20;
  vector<long long> v;
  long long set, origsize = 0, redsize = 0;
  void push(long long val) {
    origsize++;
    if (val) v.push_back(val);
  }
  void clear() {
    v.clear();
    set = 0, redsize = 0;
  }
  void eliminate() {
    set = redsize = 0;
    for (long long bit = 0; bit <= no_of_bits; bit++) {
      bool check = false;
      for (long long i = redsize; i < v.size(); i++) {
        if ((v[i] >> bit) & 1) {
          swap(v[i], v[redsize]);
          check = true;
          break;
        }
      }
      if (check) {
        for (long long i = redsize + 1; i < v.size(); i++) {
          if ((v[i] >> bit) & 1) v[i] ^= v[redsize];
        }
        redsize++;
      }
    }
    v.resize(redsize);
    for (auto it : v) set |= it;
  }
  Gaussian& operator=(Gaussian& orig) {
    v = orig.v;
    set = orig.set;
    redsize = orig.redsize;
    origsize = orig.origsize;
    return *this;
  }
};
long long modexpo(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b = b / 2;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, q;
  cin >> n >> q;
  Gaussian arr[n];
  val.resize(n);
  for (long long i = 0; i < n; i++) cin >> val[i];
  arr[0].push(val[0]);
  arr[0].eliminate();
  for (long long i = 1; i < n; i++) {
    arr[i] = arr[i - 1];
    arr[i].push(val[i]);
    arr[i].eliminate();
  }
  while (q--) {
    long long l, x;
    cin >> l >> x;
    Gaussian temp = arr[--l];
    temp.push(x);
    temp.eliminate();
    long long ans = 0;
    if (temp.redsize == arr[l].redsize)
      ans = modexpo(2, arr[l].origsize - arr[l].redsize);
    cout << ans << "\n";
  }
  return 0;
}
