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: Wrong Answer)

// #define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;

#define int long long
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define rep(i, n) for (int i = 0; i < n; ++i)
#define REP(i, n) for (int i = 0; i < n; ++i)
#define range(i,a,b) ((a)<=(i) && (i)<(b))
#define debug(x) cout << #x << " = " << (x) << endl;
#define fs first
#define sc second
#define pb push_back
#define eb emplace_back
#define SP << " " <<

typedef long long ll;
typedef pair<int, int> P;
typedef tuple<int, int, int> T;
typedef vector<int> vec;
typedef vector<P> pvec;
typedef vector<vector<int>> vvec;
typedef vector<vector<P>> pvvec;
typedef priority_queue<int> PQI;
typedef priority_queue<P> PQP;
typedef priority_queue<int,vector<int>,greater<int>> PQIG;
typedef priority_queue<P,vector<P>,greater<P>> PQPG;

const vector<int> DX = {0, -1, 0, 1, 1, 1, -1, -1};
const vector<int> DY = {1, 0, -1, 0, 1, -1, 1, -1};
constexpr int MOD = (1000000007);
// const int MOD = (998244353);
// const int INF = (1 << 30); // 1073741824
const ll INF = (1LL << 60); // 1152921504606846976
const double PI = (3.141592653589794);
const double EPS = (0.0000000001); // 10^(-10)

template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b; return 1;} return 0;}
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b; return 1;} return 0;}
template<class T> inline T ceil(T a, T b) {return T((a + b - 1) / b);}
template<class T> inline T round(T a, T b) {return T(a / b);}
template< typename T1, typename T2 > istream &operator>>(istream &is, pair< T1, T2 > &p) { is >> p.first >> p.second; return is; }
template <class T> inline void out(T &a) { bool flag=true; for(auto&x:a){if(flag) {flag=false;} else{ cout << ' '; } cout<<x; } cout << endl; }



//----------------------------------------------------------------
int nmax=200000; // 2*(10^5)
vvec G(nmax);





void solve4ts()
{
    
    int a,b,c,d,e;
    cin>>a>>b>>c>>d>>e;
    int x=c-a;
    x*=60;
    x+=(60+(b-d))%60;
    cout<<x-e<<endl;
    
}
//-----------------------------------------------------------------

signed main(){ ios::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(15);
    int repeat = 1;
    // cin >> repeat;
    while(repeat--) solve4ts();
}