#include <bits/stdc++.h>
using namespace std;
long long int n, q;
long long int arr[300005];
long long int dist[22][22];
long long int mindist[300005][22];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> q;
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
  for (long long int i = 0; i <= 20; i++) {
    for (long long int j = 0; j <= 20; j++) {
      if (i != j) dist[i][j] = INT_MAX;
    }
  }
  for (long long int i = n; i >= 1; i--) {
    vector<long long int> z, o;
    for (long long int j = 0; j <= 20; j++) {
      if (arr[i] & (1 << j)) {
        mindist[i][j] = 0;
        o.push_back(j);
        for (long long int k = 0; k <= 20; k++) {
          if (arr[i] & (1 << k)) dist[j][k] = 0;
        }
      } else
        z.push_back(j);
    }
    for (auto it0 : z) {
      long long int m = INT_MAX;
      for (auto it1 : o) {
        m = min(m, dist[it1][it0]);
      }
      for (auto it1 : o) {
        dist[it1][it0] = m;
      }
      mindist[i][it0] = m;
    }
    for (long long int j = 0; j <= 20; j++) {
      for (long long int k = 0; k <= 20; k++) dist[j][k]++;
    }
  }
  while (q--) {
    long long int x, y;
    cin >> x >> y;
    long long int m = INT_MAX;
    for (long long int j = 0; j <= 20; j++) {
      if (arr[y] & (1 << j)) m = min(m, mindist[x][j]);
    }
    if (m <= (y - x))
      cout << "Shi\n";
    else
      cout << "Fou\n";
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

Toad Pimple has an array of integers a_1, a_2, …, a_n.

We say that y is reachable from x if x<y and there exists an integer array p such that x = p_1 < p_2 < … < p_k=y, and a_{p_i}  \&  a_{p_{i+1}} > 0 for all integers i such that 1 ≤ i < k.

Here \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

You are given q pairs of indices, check reachability for each of them.

Input

The first line contains two integers n and q (2 ≤ n ≤ 300 000, 1 ≤ q ≤ 300 000) — the number of integers in the array and the number of queries you need to answer.

The second line contains n space-separated integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 300 000) — the given array.

The next q lines contain two integers each. The i-th of them contains two space-separated integers x_i and y_i (1 ≤ x_i < y_i ≤ n). You need to check if y_i is reachable from x_i. 

Output

Output q lines. In the i-th of them print "Shi" if y_i is reachable from x_i, otherwise, print "Fou".

Example

Input


5 3
1 3 0 2 1
1 3
2 4
1 4


Output


Fou
Shi
Shi

Note

In the first example, a_3 = 0. You can't reach it, because AND with it is always zero. a_2  \&  a_4 > 0, so 4 is reachable from 2, and to go from 1 to 4 you can use p = [1, 2, 4].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int n, q;
long long int arr[300005];
long long int dist[22][22];
long long int mindist[300005][22];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> q;
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
  for (long long int i = 0; i <= 20; i++) {
    for (long long int j = 0; j <= 20; j++) {
      if (i != j) dist[i][j] = INT_MAX;
    }
  }
  for (long long int i = n; i >= 1; i--) {
    vector<long long int> z, o;
    for (long long int j = 0; j <= 20; j++) {
      if (arr[i] & (1 << j)) {
        mindist[i][j] = 0;
        o.push_back(j);
        for (long long int k = 0; k <= 20; k++) {
          if (arr[i] & (1 << k)) dist[j][k] = 0;
        }
      } else
        z.push_back(j);
    }
    for (auto it0 : z) {
      long long int m = INT_MAX;
      for (auto it1 : o) {
        m = min(m, dist[it1][it0]);
      }
      for (auto it1 : o) {
        dist[it1][it0] = m;
      }
      mindist[i][it0] = m;
    }
    for (long long int j = 0; j <= 20; j++) {
      for (long long int k = 0; k <= 20; k++) dist[j][k]++;
    }
  }
  while (q--) {
    long long int x, y;
    cin >> x >> y;
    long long int m = INT_MAX;
    for (long long int j = 0; j <= 20; j++) {
      if (arr[y] & (1 << j)) m = min(m, mindist[x][j]);
    }
    if (m <= (y - x))
      cout << "Shi\n";
    else
      cout << "Fou\n";
  }
  return 0;
}

```