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 N = 1e5 + 10;
const int X = (1 << 20);
const int D = 20;
const int MOD = 1e9 + 7;
int n, q, a[N];
vector<pair<int, int>> qry[N];
int pw2[X], msb[X];
void build() {
  pw2[0] = 1;
  for (int i = 1; i < X; ++i) pw2[i] = (2 * pw2[i - 1]) % MOD;
  msb[0] = -1;
  for (int i = 1; i < X; ++i) msb[i] = msb[i >> 1] + 1;
}
set<int, greater<int>> basis;
inline void add(int x) {
  for (int b : basis) {
    if (msb[b] == msb[x]) x ^= b;
  }
  if (x != 0) basis.insert(x);
}
inline bool chk(int x) {
  for (int b : basis) {
    if (msb[b] == msb[x]) x ^= b;
  }
  return x == 0;
}
int ans[N];
int main() {
  build();
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= q; ++i) {
    int l, x;
    scanf("%d%d", &l, &x);
    qry[l].push_back({x, i});
  }
  for (int i = 1; i <= n; ++i) {
    add(a[i]);
    for (pair<int, int> p : qry[i]) {
      int x = p.first;
      int j = p.second;
      if (chk(x)) ans[j] = pw2[i - basis.size()];
    }
  }
  for (int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
}
