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 = (1 << 20) + 10;
const int MOD = 1e9 + 7;
int n, q, a[MAXN], l[MAXN], x[MAXN], sec[MAXN];
int par[MAXN], tt[MAXN], t2 = 1;
int save[MAXN];
vector<int> bases;
int ans[MAXN];
int getPar(int v) { return par[v] == v ? v : par[v] = getPar(par[v]); }
bool cmp(int q1, int q2) { return l[q1] < l[q2]; }
void addBase(int base) {
  bases.push_back(base);
  for (int i = 0; i < 1 << 20; i++) tt[i] = 1ll * tt[i] * t2 % MOD;
  t2 = 1;
  for (int i = 0; i < 1 << 20; i++) {
    int pu = getPar(i), pv = getPar(i ^ base);
    if (pu == pv) continue;
    par[pu] = pv;
    tt[pv] += tt[pu];
    if (tt[pv] >= MOD) tt[pv] -= MOD;
  }
}
void addVec(int ind) {
  int v = a[ind];
  for (int u : bases)
    if (v >> save[u] & 1) v ^= u;
  if (v)
    addBase(v);
  else {
    t2 += t2;
    if (t2 >= MOD) t2 -= MOD;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 0; i < MAXN; i++) save[i] = 31 - __builtin_clz(i);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < q; i++) cin >> l[i] >> x[i];
  iota(sec, sec + q, 0);
  sort(sec, sec + q, cmp);
  iota(par, par + MAXN, 0);
  tt[0] = 1;
  int cur = 0;
  for (int i = 0; i < q; i++) {
    int id = sec[i];
    while (cur < l[id]) addVec(cur++);
    ans[id] = 1ll * tt[getPar(x[id])] * t2 % MOD;
  }
  for (int i = 0; i < q; i++) cout << ans[i] << "\n";
  return 0;
}
