#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const long long INF = std::numeric_limits<long long>::max();
const int MX = 100101;
const long double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270L;
template <typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) {
  return pair<T, T>(a.first + b.first, a.second + b.second);
}
template <typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) {
  return pair<T, T>(a.first - b.first, a.second - b.second);
}
template <typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) {
  return (a.first * b.first + a.second * b.second);
}
template <typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) {
  return (a.first * b.second - a.second * b.first);
}
const int lg = 20;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int N, Q;
  cin >> N >> Q;
  vector<int> sk(N);
  for (int i = 0; i < N; ++i) cin >> sk[i];
  vector<vector<int> > last(N + 1, vector<int>(lg, MOD));
  for (int i = 0; i < lg; ++i) {
    int curr = MOD;
    for (int j = N - 1; j >= 0; --j) {
      last[j][i] = curr;
      if (sk[j] & (1 << i)) curr = j;
    }
  }
  vector<vector<int> > dp(N + 1, vector<int>(lg, MOD));
  for (int i = 0; i < lg; ++i) {
    for (int j = N - 1; j >= 0; j--) {
      dp[j][i] = MOD;
      if (sk[j] & (1 << i)) {
        dp[j][i] = j;
        continue;
      }
      for (int k = 0; k < lg; ++k) {
        if (sk[j] & (1 << k) and last[j][k] != MOD)
          dp[j][i] = min(dp[j][i], dp[last[j][k]][i]);
      }
    }
  }
  int a, b;
  for (int i = 0; i < Q; ++i) {
    cin >> a >> b;
    a--;
    b--;
    bool good = false;
    for (int j = 0; j < lg; ++j) {
      if (sk[b] & (1 << j) and dp[a][j] <= b) {
        good = true;
        break;
      }
    }
    printf(good ? "Shi\n" : "Fou\n");
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int MOD = 1000000007;
const long long INF = std::numeric_limits<long long>::max();
const int MX = 100101;
const long double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270L;
template <typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) {
  return pair<T, T>(a.first + b.first, a.second + b.second);
}
template <typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) {
  return pair<T, T>(a.first - b.first, a.second - b.second);
}
template <typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) {
  return (a.first * b.first + a.second * b.second);
}
template <typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) {
  return (a.first * b.second - a.second * b.first);
}
const int lg = 20;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int N, Q;
  cin >> N >> Q;
  vector<int> sk(N);
  for (int i = 0; i < N; ++i) cin >> sk[i];
  vector<vector<int> > last(N + 1, vector<int>(lg, MOD));
  for (int i = 0; i < lg; ++i) {
    int curr = MOD;
    for (int j = N - 1; j >= 0; --j) {
      last[j][i] = curr;
      if (sk[j] & (1 << i)) curr = j;
    }
  }
  vector<vector<int> > dp(N + 1, vector<int>(lg, MOD));
  for (int i = 0; i < lg; ++i) {
    for (int j = N - 1; j >= 0; j--) {
      dp[j][i] = MOD;
      if (sk[j] & (1 << i)) {
        dp[j][i] = j;
        continue;
      }
      for (int k = 0; k < lg; ++k) {
        if (sk[j] & (1 << k) and last[j][k] != MOD)
          dp[j][i] = min(dp[j][i], dp[last[j][k]][i]);
      }
    }
  }
  int a, b;
  for (int i = 0; i < Q; ++i) {
    cin >> a >> b;
    a--;
    b--;
    bool good = false;
    for (int j = 0; j < lg; ++j) {
      if (sk[b] & (1 << j) and dp[a][j] <= b) {
        good = true;
        break;
      }
    }
    printf(good ? "Shi\n" : "Fou\n");
  }
}

```