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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const int LOG = 20;
int basis[LOG];
int rk;
void insert(int mask) {
  for (int i = LOG - 1; i >= 0; i--)
    if (mask >> i & 1) {
      if (!basis[i]) {
        basis[i] = mask, rk++;
        return;
      }
      mask ^= basis[i];
    }
}
int get(int mask) {
  int ret = 0;
  for (int i = LOG - 1; i >= 0; i--)
    if (mask >> i & 1) {
      if (!basis[i]) return -1;
      mask ^= basis[i], ret |= (1 << i);
    }
  return ret;
}
const int MOD = 1e9 + 7;
long long expo(long long a, long long b) {
  if (!b) return 1;
  long long ans = expo(a * a % MOD, b / 2);
  return b % 2 ? ans * a % MOD : ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<int> v(n);
  for (int& i : v) cin >> i;
  vector<pair<int, pair<int, int>>> qu;
  for (int i = 0; i < q; i++) {
    int l, x;
    cin >> l >> x;
    l--;
    qu.push_back({l, {x, i}});
  }
  sort(qu.begin(), qu.end());
  vector<long long> ans(q);
  int at = -1;
  for (auto i : qu) {
    while (at < i.first) insert(v[++at]);
    if (get(i.second.first) + 1)
      ans[i.second.second] = expo(2, at + 1 - rk);
    else
      ans[i.second.second] = 0;
  }
  for (int i : ans) cout << i << '\n';
  exit(0);
}
