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)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MOD1 1000000007
#define MOD2 998244353
#define LIMIT1 200002
#define LIMIT2 500002
typedef long long ll;
typedef long double ld;
typedef const void cv;
#define rep(i,n) for((i)=0;(i)<(n);(i)++)
#define per(i,n) for((i)=(n)-1;(i)>=0;(i)--)
#define max(a,b) ((a)>(b) ? (a) : (b))
#define min(a,b) ((a)<(b) ? (a) : (b))
#define zt(a,b) (max((a),(b))-min((a),(b)))
#define absl(v) labs((v))
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll nPr(int n,int r){ll i,result=1;rep(i,r){result*=(ll)(n-i);}return result;}
ll nCr(int n,int r){ll i,result=1;if(n<r) return 0;rep(i,min(r,n-r)){result*=(ll)(n-i);result/=(ll)(i+1);}return result;}
#define nHr(n,r) nCr((int)((n)+(r)+1),(int)(r))
#define sankaku(x) (((x)*((x)+1))/2)
int dx[8]={1,0,-1,0,1,-1,-1,1},dy[8]={0,1,0,-1,1,1,-1,-1};
int upint(cv *a, cv *b) { return *(int *)a < *(int *)b ? -1 : *(int *)a > *(int *)b ? 1 : 0; }
int downint(cv *a, cv *b) { return *(int *)a < *(int *)b ? 1 : *(int *)a > *(int *)b ? -1 : 0; }
int upchar(cv *left, cv *right) {return strcmp((char *)left,(char *)right);}
int downchar(cv *left, cv *right) {return strcmp((char *)right,(char *)left);}
void tmpInit(int *c,ll m) {ll i;rep(i,m) c[i]=i;}
ll modpow(ll a, ll n, ll mod) {
    ll res = 1;
    while (n > 0) {
        if (n & 1){
            res *= a; if(mod>=1) res %= mod;
        }
        a *= a; if(mod>=1) a %= mod;
        n >>= 1;
    }
    return res;
}
void initialize(){
    
}
int main(void){
  initialize();
  ll n,i,j,k,result=0,h[2],m[2];
  int a[LIMIT2]={0};
  char s[LIMIT1];
  
  scanf("%lld %lld %lld %lld %lld",&h[0],&m[0],&h[1],&m[1],&k);
  result = h[1]*60+m[1];
  result -= h[0]*60-m[0];
  result -= k;
  printf("%lld",result);
  
  return 0;
}