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>
#pragma GCC optimize("Ofast")
using namespace std;
const long double eps = 1e-7;
const int inf = 1000000010;
const long long INF = 10000000000000010LL;
const int mod = 1000000007;
const int MAXN = 100010, LOG = 20;
int n, q, u, v, x, y, t = 1, a, b;
int A[MAXN];
int ans[MAXN];
int lb[LOG];
vector<pair<int, int> > query[MAXN];
void update(int x) {
  for (int i = 0; i < LOG; i++)
    if (x & (1 << i)) x ^= lb[i];
  if (!x) {
    t = (t << 1) % mod;
    return;
  }
  lb[__builtin_ctz(x)] = x;
}
int ok(int x) {
  for (int i = 0; i < LOG; i++)
    if (x & (1 << i)) x ^= lb[i];
  if (!x) return t;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> A[i];
  for (int i = 1; i <= q; i++) {
    cin >> x >> y;
    query[x].push_back({i, y});
  }
  for (int i = 1; i <= n; i++) {
    update(A[i]);
    for (pair<int, int> p : query[i]) ans[p.first] = ok(p.second);
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << '\n';
  return 0;
}
