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 long long N = 2e6 + 10;
const long long oo = 1e9;
const long long MOD = 1e9 + 7;
long long n, q, sum = 1, pointer = 0;
long long a[N], ans[N];
bool kt[N];
vector<long long> pt;
struct query {
  long long l, x, id;
} qu[N];
bool cmp(const query x, const query y) { return x.l < y.l; }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; ++i) cin >> a[i];
  for (long long i = 1; i <= q; ++i) cin >> qu[i].l >> qu[i].x, qu[i].id = i;
  sort(qu + 1, qu + 1 + q, cmp);
  pt.push_back(0);
  kt[0] = true;
  for (long long i = 1; i <= q; ++i) {
    while (pointer < qu[i].l) {
      pointer++;
      if (!kt[a[pointer]]) {
        long long las = pt.size();
        kt[a[pointer]] = true;
        pt.push_back(a[pointer]);
        for (long long j = 0; j <= las - 1; ++j)
          if (!kt[a[pointer] ^ pt[j]]) {
            kt[a[pointer] ^ pt[j]] = true;
            pt.push_back(a[pointer] ^ pt[j]);
          }
      } else
        sum = (sum * 2) % MOD;
    }
    if (kt[qu[i].x])
      ans[qu[i].id] = sum;
    else
      ans[qu[i].id] = 0;
  }
  for (long long i = 1; i <= q; ++i) cout << ans[i] << '\n';
}
