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 + 5, mod = 1e9 + 7;
int N, basis[30], Q, sz;
int a[maxn], res[maxn];
vector<pair<int, int> > query[maxn];
int pow2[maxn];
signed main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  function<void(int)> ins = [&](int mask) {
    for (int i = 29; i >= 0; --i)
      if (mask >> i & 1) {
        if (!basis[i]) {
          basis[i] = mask;
          ++sz;
          return;
        } else {
          mask ^= basis[i];
        }
      }
  };
  function<bool(int)> have = [&](int mask) {
    for (int i = 29; i >= 0; --i)
      if (mask >> i & 1) {
        if (!basis[i]) return false;
        mask ^= basis[i];
      }
    return true;
  };
  cin >> N >> Q;
  pow2[0] = 1;
  for (int i = 1; i <= N; ++i) {
    cin >> a[i];
    pow2[i] = 2ll * pow2[i - 1] % mod;
  }
  for (int i = 1; i <= Q; ++i) {
    int p, v;
    cin >> p >> v;
    query[p].emplace_back(v, i);
  }
  for (int i = 1; i <= N; ++i) {
    ins(a[i]);
    for (auto& all : query[i]) {
      if (have(all.first)) res[all.second] = pow2[i - sz];
    }
  }
  for (int i = 1; i <= Q; ++i) cout << res[i] << '\n';
}
