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 = 1e5 + 10;
const int bits = 21;
const int lim = 1e7;
const long long mod = 1e9 + 7;
struct query {
  int l, x, id;
  long long ans;
};
int n, q;
int a[maxn];
query w[maxn];
int base[bits];
long long curAns = 1;
void add(int x) {
  int b = x;
  for (int i = 0; i < bits; ++i) {
    if (b >> i & 1) b ^= base[i];
  }
  if (!b) {
    curAns *= 2;
    curAns %= mod;
  } else {
    int p = 0;
    while (!(b >> p & 1)) p++;
    for (int i = 0; i < bits; ++i) {
      if (base[i] >> p & 1) base[i] ^= b;
    }
    base[p] = b;
  }
}
long long cnt(int x) {
  for (int i = 0; i < bits; ++i) {
    if (x >> i & 1) x ^= base[i];
  }
  if (x)
    return 0;
  else
    return curAns;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < q; ++i) {
    cin >> w[i].l >> w[i].x;
    w[i].l--;
    w[i].id = i;
  }
  sort(w, w + q, [](query a, query b) { return a.l < b.l; });
  int cur = 0;
  for (int i = 0; i < q; ++i) {
    while (cur <= w[i].l) add(a[cur++]);
    w[i].ans = cnt(w[i].x);
  }
  sort(w, w + q, [](query a, query b) { return a.id < b.id; });
  for (int i = 0; i < q; ++i) printf("%lld\n", w[i].ans);
}
