Problem:
Score : 100 points

Problem Statement

Given are three integers A_1, A_2, and A_3.

If A_1+A_2+A_3 is greater than or equal to 22, print bust; otherwise, print win.

Constraints

1 \leq A_i \leq 13 \ \ (i=1,2,3)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A_1 A_2 A_3

Output

If A_1+A_2+A_3 is greater than or equal to 22, print bust; otherwise, print win.

Sample Input 1

5 7 9

Sample Output 1

win

5+7+9=21, so print win.

Sample Input 2

13 7 2

Sample Output 2

bust

13+7+2=22, so print bust.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<int> vi;
typedef vector<long long int> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long int>> vvl;
typedef pair<int, int> pii;
typedef pair<long long int, long long int> pll;
typedef vector<pair<int, int>> vii;
typedef vector<pair<long long int, long long int>> vll;
#define fast() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define test() long long int t; cin >> t; cin.ignore(); while(t--) 
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define mp make_pair
#define MOD 1000000007
#define SMOD 998244353
#define pq priority_queue
#define all(x) x.begin(),x.end()
#define SP(x) setprecision(x)
#define sp " "
#define inp(a) ll a;cin>>a;cin.ignore()
#define inps(a,b) ll a,b;cin>>a>>b;cin.ignore()
#define out(a) cout<<SP(20)<<a<<'\n'
#define outs(a,b) cout<<SP(20)<<a<<" "<<b<<'\n'
#define ins insert
#define er erase
#define lb lower_bound
#define ub upper_bound
#define printclock cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";
#define inpa(a,n) ll a[n]; for(int i=0; i<n; i++){cin>>a[i];}
#define inpv(a,n) vl a; for(int i=0; i<n; i++){inp(x);a.pb(x);}
#define outa(a,n) for(int i=0;i<n;i++){cout<<a[i]<<" ";}cout<<'\n';
#define outv(v) for(auto i:v){cout<<i<<" ";}cout<<'\n';
set<char> vow = {'a','e','i','o','u','A','E','I','O','U'};
ll xpow(ll n, ll m){if(m==0){return 1;}else{ll x = xpow(n, m/2);if(m&1){return x*x*n;}else{return x*x;}}}
ll xmod(ll a, ll m){return (((a%m)+m)%m);}
ll xmodpow(ll n, ll m, ll y){if(m==0){return 1;}else{ll x = xmodpow(n, m/2, y);if(m&1){return xmod((xmod((xmod(x,y)*xmod(x,y)),y)*xmod(n,y)),y);}else{return xmod((xmod(x,y)*xmod(x,y)),y);}}}
 
int main(){
    fast();
 
/* you get what you work for not what you wish for */
    
    inpv(a, 3);
    ll sum = 0;
    for(auto i:a){
        sum+=i;
    }
    if(sum>=22){
        cout << "bust" << endl;
    }
    else{
        cout << "win" << endl;
    }
}