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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
int n, q;
vector<pair<int, int> > Q[100009];
vector<int> ans(100009, 0), Powers(100009, 0), A(100009, 0);
int main() {
  Powers[0] = 1;
  for (int i = 1; i < 100009; i++) Powers[i] = Powers[i - 1] * 2 % 1000000007;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  for (int i = 0; i < q; i++) {
    int l, x;
    scanf("%d%d", &l, &x);
    Q[l].push_back(pair<int, int>(x, i));
  }
  int f[20];
  memset(f, -1, sizeof f);
  int z = 0;
  for (int i = 1; i <= n; i++) {
    int a = A[i];
    int ok = -1;
    for (int j = 19; j >= 0; j--) {
      if (1 << j & a) {
        if (f[j] == -1) {
          ok = j;
          break;
        } else
          a ^= f[j];
      }
    }
    if (ok == -1)
      ++z;
    else
      f[ok] = a;
    for (auto e : Q[i]) {
      int x = e.first;
      for (int j = 19; j >= 0; j--) {
        if (1 << j & x) {
          if (f[j] == -1) break;
          x ^= f[j];
        }
      }
      ans[e.second] = (x == 0 ? Powers[z] : 0);
    }
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
}
