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"

using namespace std;

#define rep(i,n) for(int i=0;i<(int)(n);i++)
void sort2array(double *a,long *b,long n){
    for(int i=0;i<n;i++){
        b[i]=i;
    }
    sort(b,b+n,[a](long i,long j)->bool {return a[i]<a[j];} );
}
//bを何回足せばaを超えるか
int wtover(int a,int b){
    if(a%b>0)return (a/b)+1;
    else return a/b;
}
const int INF=100000001,ID_MAX=20;
int bi_e[ID_MAX]={0};
//2進数表示したときの最高桁
int bi_max(long n){
    int m=0;
    for(m;pow(2,m)<=n;m++);m=m-1;
    return m;
}
//bi_eに二進数表示したやつを代入
void bi_exs(long n){
    rep(i,ID_MAX){
        bi_e[i]=0;
    }
    if(n<pow(2,ID_MAX)){
        int m=bi_max(n);
        long l;
        for(m;m>-1;m--){
            l=pow(2,m);
            bi_e[m]=n/l;
            n=n%l;
        }
    }
}
//x^n mod m (nが負の時は０)
long myPow(long x, long n, long m){
  if(n<0)return 0;
  if(n == 0)
    return 1;
  if(n % 2 == 0)
    return myPow(x * x % m, n / 2, m);
  else
    return x * myPow(x, n - 1, m) % m;
}


int main(){
    int i,s,w;
    //入力
    cin>>s>>w;
    //処理
    //出力
    w>=s?cout<<"unsafe"<<endl:cout<<"safe"<<endl;
}