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 somod = 1e9 + 7;
const long long maxn = 1e5 + 7;
long long n, q, basis[21], a[maxn], sz, ans[maxn], Pow[maxn];
vector<pair<long long, long long>> vt[maxn];
long long GetBit(const long long &a, const long long &b) {
  return (a >> b & 1);
}
void Add(long long x) {
  for (long long i = 0; i < 20; i++) {
    if (GetBit(x, i)) {
      if (!basis[i]) {
        basis[i] = x;
        sz++;
        return;
      }
      x ^= basis[i];
    }
  }
}
void Solve(long long id, long long pos, long long val) {
  for (long long i = 0; i < 20; i++) {
    if (GetBit(val, i)) val ^= basis[i];
  }
  if (val)
    ans[id] = 0;
  else
    ans[id] = Pow[pos - sz];
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  Pow[0] = 1;
  for (long long i = 1; i <= n; i++) {
    Pow[i] = Pow[i - 1] * 2;
    if (Pow[i] >= somod) Pow[i] -= somod;
  }
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= q; i++) {
    long long a, b;
    cin >> a >> b;
    vt[a].push_back({i, b});
  }
  for (long long i = 1; i <= n; i++) {
    Add(a[i]);
    for (auto j : vt[i]) {
      Solve(j.first, i, j.second);
    }
  }
  for (long long i = 1; i <= q; i++) cout << ans[i] << '\n';
}
