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 MOD = (int)1e9 + 7, MAXN = 100005, LOGN = 18, MAXM = 5005;
inline void S(int& n) {
  n = 0;
  int ch = getchar();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = (n << 3) + (n << 1) + ch - '0', ch = getchar();
  }
  n = n * sign;
}
const int SHIFT = 30, ETF = MOD - 1, LIM = (int)1e9;
const long long int INF = (long long int)1e18 + 1;
const double PI = (double)acos(-1), EPSILON = 1e-10;
set<int> present;
vector<pair<int, pair<int, int> > > queries;
void pre() { present.insert(0); }
int ans[MAXN];
int main() {
  present.insert(0);
  int n, q;
  S(n);
  S(q);
  queries.resize(q);
  int arr[n + 1];
  for (int i = 1; i < n + 1; i++) {
    S(arr[i]);
  }
  for (int i = 0; i < q; i++) {
    S(queries[i].first);
    S(queries[i].second.first);
    queries[i].second.second = i;
  }
  sort(queries.begin(), queries.end());
  int ptr = 0;
  int cur_ans = 1;
  for (int i = 1; i < n + 1; i++) {
    if (present.find(arr[i]) == present.end()) {
      vector<int> newadd;
      for (__typeof((present).begin()) itr = (present).begin();
           itr != (present).end(); itr++) {
        int num = *itr;
        newadd.push_back(num ^ arr[i]);
      }
      for (int j = 0; j < newadd.size(); j++) {
        present.insert(newadd[j]);
      }
    } else {
      cur_ans = (cur_ans + cur_ans) < 0
                    ? cur_ans + cur_ans + MOD
                    : ((cur_ans + cur_ans >= MOD) ? cur_ans + cur_ans - MOD
                                                  : cur_ans + cur_ans);
    }
    while (queries[ptr].first == i) {
      int qn = queries[ptr].second.second;
      int x = queries[ptr].second.first;
      if (present.find(x) != present.end()) {
        ans[qn] = cur_ans;
      } else {
        ans[qn] = 0;
      }
      ptr++;
    }
  }
  for (int i = 0; i < q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
