Problem:
Score : 100 points

Problem Statement

In this problem, we use the 24-hour clock.

Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.)
He has decided to study for K consecutive minutes while he is up.
What is the length of the period in which he can start studying?

Constraints

0 \le H_1, H_2 \le 23

0 \le M_1, M_2 \le 59

The time H_1 : M_1 comes before the time H_2 : M_2.

K \ge 1

Takahashi is up for at least K minutes.

All values in input are integers (without leading zeros).

Input

Input is given from Standard Input in the following format:

H_1 M_1 H_2 M_2 K

Output

Print the length of the period in which he can start studying, as an integer.

Sample Input 1

10 0 15 0 30

Sample Output 1

270

Takahashi gets up at exactly ten in the morning and goes to bed at exactly three in the afternoon.
It takes 30 minutes to do the study, so he can start it in the period between ten o'clock and half-past two. The length of this period is 270 minutes, so we should print 270.

Sample Input 2

10 0 12 0 120

Sample Output 2

0

Takahashi gets up at exactly ten in the morning and goes to bed at exactly noon. It takes 120 minutes to do the study, so he has to start it at exactly ten o'clock. Thus, we should print 0.

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

#include <bits/stdc++.h>
 
#define int         long long
#define uint        unsigned int
#define ld          long double
#define showoff     ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define pb          push_back
#define pii         pair<int,int>
#define FOR(i,a,b)  for(int i=a;i<b;++i)
#define RFOR(i,a,b) for(int i=a;i>b;--i)
#define f           first
#define se          second
#define maxn        200005
#define all(v)      v.begin(),v.end()
#define sz(x)       (int)x.size()
#define mod         998244353
#define pqueue      priority_queue<pii>
#define pdqueue     priority_queue< int,vector<int> ,greater< int >>
 
using namespace std;
 
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  
int power(int a,int n)
{
    a %= mod;
    if(n == 1)return a;
    if(n == 0)return 1;
    if(n%2)return (a*(power((a*a)%mod,n/2)%mod))%mod;
    return power((a*a)%mod,n/2)%mod;
}
 
const int inf = (int) 1e18;
 
int inverse(int x){
    return power(x,mod-2)%mod;//little fermat....
}  

void solve(){
    int h1,m1,h2,m2,k;
    cin >> h1 >> m1 >> h2 >> m2 >> k;
    int tot = 0;
    if(h2 > h1)tot = m2+(h2-h1-1)*60+60-m1;
    else tot = m2-m1;
    cout << max(0LL,tot-k) << "\n";
}       

 
signed main()   
{
    showoff;
    int T = 1;
    //cin >> T;
    FOR(t,1,T+1){
        solve();
        cout << "\n";
    }
    return 0;
}
//*->for large size of matrix take int not long long if possible......
//*->always take maximum as inf for safer side ...