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;
int const MAX = 1e5 + 41;
int const MOD = 1e9 + 7;
int add(int a, int b) { return (a + b) % MOD; }
int n, q;
int a[MAX];
int l[MAX];
int x[MAX];
int ans[MAX];
int const B = 30;
int mask[B];
int p2[MAX];
int r;
vector<pair<int, int> > e[MAX];
int const ADD = 0;
int const ASK = 1;
bool bit(int m, int i) { return (m & (1 << i)); }
void push(int m) {
  for (int i = B - 1; i >= 0; i--) {
    if (!bit(m, i)) continue;
    if (mask[i] == -1) {
      r++;
      mask[i] = m;
      return;
    } else {
      m ^= mask[i];
    }
  }
  assert(m == 0);
}
int can(int m) {
  for (int i = B - 1; i >= 0; i--) {
    if (!bit(m, i)) continue;
    if (mask[i] == -1) return false;
    m ^= mask[i];
  }
  return true;
}
void solve() {
  p2[0] = 1;
  for (int i = 1; i <= MAX - 1; i++) {
    p2[i] = add(p2[i - 1], p2[i - 1]);
  }
  for (int i = 1; i <= n; i++) {
    e[i].push_back(make_pair(ADD, i));
  }
  for (int i = 1; i <= q; i++) {
    e[l[i]].push_back(make_pair(ASK, i));
  }
  memset(mask, 255, sizeof(mask));
  for (int i = 1; i <= n; i++) {
    for (pair<int, int> z : e[i]) {
      int t = z.first;
      int x = z.second;
      if (t == ADD) {
        push(a[x]);
      } else {
        if (can(::x[x])) {
          ans[x] = p2[i - r];
        }
      }
    }
  }
  for (int i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d %d", &l[i], &x[i]);
  }
  solve();
  return 0;
}
