Problem:
Score : 100 points

Problem Statement

There are S sheep and W wolves.

If the number of wolves is greater than or equal to that of sheep, the wolves will attack the sheep.

If the wolves will attack the sheep, print unsafe; otherwise, print safe.

Constraints

1 \leq S \leq 100

1 \leq W \leq 100

Input

Input is given from Standard Input in the following format:

S W

Output

If the wolves will attack the sheep, print unsafe; otherwise, print safe.

Sample Input 1

4 5

Sample Output 1

unsafe

There are four sheep and five wolves. The number of wolves is not less than that of sheep, so they will attack them.

Sample Input 2

100 2

Sample Output 2

safe

Many a sheep drive away two wolves.

Sample Input 3

10 10

Sample Output 3

unsafe

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
#define ss  ios_base::sync_with_stdio(false);cin.tie(NULL);
#define present(container, element) (container.find(element) != container.end())
#define ss  ios_base::sync_with_stdio(false);cin.tie(NULL);
#define all(c) c.begin(), c.end()
#define pb push_back
#define mp make_pair
#define mod (1000000000 + 7)
#define inf (1e10+9)
#define fi first
#define se second
#define MAXN   1000001 
typedef long long ll;
using namespace std;
using namespace __gnu_pbds; 
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> new_set;

long long fastpow(long long n, long long x, long long m){
	
    long long ret = 1;
    
    while(x){
    if(x&1) (ret*=n)%=m;
    x>>=1;
    (n*=n)%=m;
    }
  
 return ret;
}
ll add(ll a,ll b){
	return((a%mod)+(b%mod))%mod;
}
ll prod(ll a,ll b){
	a%=mod;
	b%=mod;
	return ((a%mod)*(b%mod)%mod);
}
int main()
{
	ss
    ll a,b;
    cin>>a>>b;
    if(b>=a){
    	cout<<"unsafe\n";
	}
	else cout<<"safe\n";
    
    
    
    
    
   
    return  0;
}