Codeforces 1169E 超难

E. And Reachability

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Toad Pimple has an array of integers a1,a2,…,ana1,a2,…,an.

We say that yy is reachable from xx if x<yx<y and there exists an integer array pp such that x=p1<p2<…<pk=yx=p1<p2<…<pk=y, and api&api+1>0api&api+1>0 for all integers ii such that 1≤i<k1≤i<k.

Here && denotes the bitwise AND operation.

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

Input

The first line contains two integers nn and qq (2≤n≤3000002≤n≤300000, 1≤q≤3000001≤q≤300000) — the number of integers in the array and the number of queries you need to answer.

The second line contains nn space-separated integers a1,a2,…,ana1,a2,…,an (0≤ai≤3000000≤ai≤300000) — the given array.

The next qq lines contain two integers each. The ii-th of them contains two space-separated integers xixi and yiyi (1≤xi<yi≤n1≤xi<yi≤n). You need to check if yiyi is reachable from xixi.

Output

Output qq lines. In the ii-th of them print "Shi" if yiyi is reachable from xixi, otherwise, print "Fou".

Example

input

Copy

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

output

Copy

Fou
Shi
Shi

Note

In the first example, a3=0a3=0. You can't reach it, because AND with it is always zero. a2&a4>0a2&a4>0, so 44 is reachable from 22, and to go from 11 to 44 you can use p=[1,2,4]p=[1,2,4].

#include<iostream>
#include<cstdio>
#define maxn 300010
using namespace std;
int n,q,a[maxn],dp[maxn][30],las[30];
int main(){
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    for(int i=0;i<20;i++){
        las[i]=n+1;
        dp[n+1][i]=n+1;
    }
    for(int i=n;i>=1;i--){
        for(int j=0;j<20;j++)
            dp[i][j]=n+1;
        for(int j=0;j<20;j++){
            if(a[i]&(1<<j)){
                for(int k=0;k<20;k++)
                    dp[i][k]=min(dp[i][k],dp[las[j]][k]);
                las[j]=i;
                dp[i][j]=i;
            }
        }
    }
    int x,y;
    while(q--){
        cin>>x>>y;
        bool f=0;
        for(int i=0;i<20;i++){
            if(a[y]&(1<<i)&&dp[x][i]<=y){
                f=1;
                break;
            }
        }
        if(f)puts("Shi");
        else puts("Fou");
    }
    return 0;
}