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>
const long long MAX_N = 1e6 + 5;
const long long MOD = 1e9 + 7;
using namespace std;
long long a[MAX_N];
long long dp[MAX_N];
int n, q;
map<pair<int, int>, long long> answer;
vector<int> qs[MAX_N];
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  vector<pair<int, int>> queries;
  for (int i = 0; i < q; ++i) {
    int l, x;
    cin >> l >> x;
    queries.push_back({l, x});
    qs[l].push_back(x);
  }
  dp[0]++;
  set<int> st;
  st.insert(0);
  for (int i = 1; i <= n; ++i) {
    dp[i] = dp[i - 1];
    if (st.count(a[i]))
      dp[i] += dp[i - 1];
    else {
      set<int> pm = st;
      for (auto x : pm) st.insert(x ^ a[i]);
    }
    dp[i] %= MOD;
    for (auto x : qs[i]) {
      if (!st.count(x))
        answer[{i, x}] = 0;
      else
        answer[{i, x}] = dp[i];
    }
  }
  for (auto x : queries) printf("%I64d\n", answer[x]);
  return 0;
}
