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 inf = 1 << 30;
const long long INF = 1LL << 60;
const int MaxN = 100002;
const int MOD = 1e9 + 7;
int n, Q;
int bin[MaxN + 5], a[MaxN + 5];
int basis[MaxN + 5][25];
int cnt[MaxN + 5];
void Solve() {
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 20; j++) basis[i][j] = basis[i - 1][j];
    cnt[i] = cnt[i - 1];
    int cur = a[i];
    for (int j = 20; j >= 0; j--) {
      if (cur & (1 << j)) {
        if (!basis[i][j]) {
          basis[i][j] = cur;
          cnt[i]++;
          break;
        } else
          cur = cur ^ basis[i][j];
      }
    }
  }
}
bool check(int l, int x) {
  for (int i = 20; i >= 0; i--)
    if (x & (1 << i)) x = x ^ basis[l][i];
  if (x == 0)
    return true;
  else
    return false;
}
int main() {
  bin[0] = 1;
  for (int i = 1; i <= MaxN; i++) bin[i] = (bin[i - 1] * 2) % MOD;
  while (scanf("%d %d", &n, &Q) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    Solve();
    while (Q--) {
      int l, x;
      scanf("%d %d", &l, &x);
      if (!check(l, x))
        printf("0\n");
      else
        printf("%d\n", bin[l - cnt[l]]);
    }
    memset(cnt, 0, sizeof(cnt));
  }
  return 0;
}
