Toad Pimple has an array of integers $a_1, a_2, \ldots, a_n$.
We say that $y$ is reachable from $x$ if $x<y$ and there exists an integer array $p$ such that $x = p_1 < p_2 < \ldots < p_k=y$, and $a_{p_i}\, \&\, a_{p_{i+1}} > 0$ for all integers $i$ such that $1 \leq i < k$.
Here $\&$ denotes the bitwise AND operation.
You are given $q$ pairs of indices, check reachability for each of them.
-----Input-----
The first line contains two integers $n$ and $q$ ($2 \leq n \leq 300\,000$, $1 \leq q \leq 300\,000$) — the number of integers in the array and the number of queries you need to answer.
The second line contains $n$ space-separated integers $a_1, a_2, \ldots, a_n$ ($0 \leq a_i \leq 300\,000$) — the given array.
The next $q$ lines contain two integers each. The $i$-th of them contains two space-separated integers $x_i$ and $y_i$ ($1 \leq x_i < y_i \leq n$). 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 $p = [1, 2, 4]$.
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop,heapify
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline

from itertools import accumulate
from functools import lru_cache

M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
 
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n')]
def li3():return [int(i) for i in input().rstrip('\n')]




n, q = li()


queue = [-1] * 20

ans = [[-1] * 20 for i in range(n + 1)]
l = li()
for i, curr in enumerate(l):

    for j in range(20):
        if curr >> j & 1:
            for k in range(20):        
                ans[i][k] = max(ans[i][k], ans[queue[j]][k])
            ans[i][j] = i
    
    for j in range(20):queue[j] = max(queue[j], ans[i][j])


queries = []
for i in range(q):queries.append(li())    
for i in range(q):
    a, b = queries[i]
    a -= 1
    b -= 1

    currans = 0

    for j in range(20):
        if (l[a] >> j) & 1 and ans[b][j] >= a:
            currans = 1
            break
    print('Shi' if currans else 'Fou')
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop,heapify
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline

from itertools import accumulate
from functools import lru_cache

M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
 
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n')]
def li3():return [int(i) for i in input().rstrip('\n')]




n, q = li()


queue = [-1] * 20

ans = [[-1] * 20 for i in range(n + 1)]
l = li()
for i, curr in enumerate(l):

    for j in range(20):
        if curr >> j & 1:
            for k in range(20):        
                ans[i][k] = max(ans[i][k], ans[queue[j]][k])
            ans[i][j] = i
    
    for j in range(20):queue[j] = max(queue[j], ans[i][j])


queries = []
for i in range(q):queries.append(li())    
for i in range(q):
    a, b = queries[i]
    a -= 1
    b -= 1

    currans = 0

    for j in range(20):
        if (l[a] >> j) & 1 and ans[b][j] >= a:
            currans = 1
            break
    print('Shi' if currans else 'Fou')