Problem:
Score: 100 points

Problem Statement

In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019.

Integers M_1, D_1, M_2, and D_2 will be given as input.

It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1.

Determine whether the date 2019-M_1-D_1 is the last day of a month.

Constraints

Both 2019-M_1-D_1 and 2019-M_2-D_2 are valid dates in the Gregorian calendar.

The date 2019-M_2-D_2 follows 2019-M_1-D_1.

Input

Input is given from Standard Input in the following format:

M_1 D_1
M_2 D_2

Output

If the date 2019-M_1-D_1 is the last day of a month, print 1; otherwise, print 0.

Sample Input 1

11 16
11 17

Sample Output 1

0

November 16 is not the last day of a month.

Sample Input 2

11 30
12 1

Sample Output 2

1

November 30 is the last day of November.

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

//IN THE NAME OF GOD;
//Test By Dev-C++5.11
#include<bits/stdc++.h>
#define ll  long long
#define isc ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define ld long double
#define ss string
#define ww while
#define vl vector<ll>
#define pb push_back
#define in insert
#define vec vector
#define itr iterator
#define gt goto
#define rep(i,n,j) for(ll i=n;i<j;i++)
#define tep(i,n,j) for(ll i=n;i>=j;i--)
#define co(n) cout << n ;
#define pll pair<ll,ll>
#define bft bool f=true
#define bff bool f=false 
using namespace std;


ll kmm(ll a, ll b) {
	return(a*b) / __gcd(a, b);
}

const ll mod = 1e9 + 7;
const ll MAXN = 2e5+5;
ll modpower(ll x,ll n)
{
    if(n==0)
        return 1;
    else if(n%2 == 0)
        return modpower((x*x)%mod,n/2);
    else
        return (x*modpower((x*x)%mod,(n-1)/2))%mod;

}


void solve(){
	ll n,x,y,ans;
	cin >> n;
	bff;
	x=n/1.08;
	rep(i,x-1,x+2 && !f){
		y=i*1.08;
		if(y==n){
			ans=i;
			f=true;
		}
	}
	if(f)
		cout << ans;
	else
		cout << ":(";
}

int main() {
	isc;
	solve();
	return 0;
}