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 = 1048576, mod = 1e9 + 7;
int a[maxn], l[maxn], x[maxn], ans[maxn];
int z[21], tpow[maxn];
vector<int> qr[maxn];
void insert(vector<int> &p, int z) {
  for (int i = 0; i < p.size(); ++i) z = min(z, z ^ p[i]);
  if (z > 0) p.push_back(z);
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 0; i < q; ++i) {
    scanf("%d %d", l + i, x + i);
    qr[l[i]].push_back(i);
  }
  tpow[0] = 1;
  for (int i = 1; i < maxn; ++i) tpow[i] = tpow[i - 1] * 1ll * 2 % mod;
  vector<int> p;
  for (int i = 1; i <= n; ++i) {
    insert(p, a[i]);
    for (int j = 0; j < qr[i].size(); ++j) {
      int t = x[qr[i][j]];
      for (int k : p) t = min(t, t ^ k);
      if (t == 0)
        ans[qr[i][j]] = tpow[i - (int)p.size()];
      else
        ans[qr[i][j]] = 0;
    }
  }
  for (int i = 0; i < q; ++i) printf("%d\n", ans[i]);
  return 0;
}
