Create a program that takes two dates as input and outputs the number of days between the two dates.

Date 1 (y1, m1, d1) is the same as or earlier than date 2 (y2, m2, d2). Date 1 is included in the number of days, not date 2. Also, take the leap year into account when calculating. The leap year conditions are as follows.

* The year is divisible by 4.
* However, a year divisible by 100 is not a leap year.
* However, a year divisible by 400 is a leap year.



Input

Given multiple datasets. The format of each dataset is as follows:


y1 m1 d1 y2 m2 d2


When any of y1, m1, d1, y2, m2, and d2 is a negative number, the input ends.

The number of datasets does not exceed 50.

Output

Output the number of days on one line for each dataset.

Example

Input

2006 9 2 2006 9 3
2006 9 2 2006 11 11
2004 1 1 2005 1 1
2000 1 1 2006 1 1
2000 1 1 2101 1 1
-1 -1 -1 -1 -1 -1


Output

1
70
366
2192
36890
#include <bits/stdc++.h>
using namespace std;

// infty
#define ULLINF (ULLONG_MAX)
#define LLINF  (LLONG_MAX)
#define IINF   (INT_MAX)
#define INF    (1<<29)

// math
#define Sq(x) ((x)*(x))

// container utility
#define ALL(x) (x).begin(), (x).end()
#define MP make_pair
#define PB push_back
#define EACH(it,c) for(__typeof((c).begin())it=(c).begin();it!=(c).end();it++)

// rep
#define REP(i,a,b) for(int i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)

// typedef
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VII;
typedef vector<VI> VVI;

typedef long long ll;

// pair util
#define FST first
#define SND second

// range validator
#define CK(n,a,b) (a<=n && n<b)

// conversion
template<class T> inline string toStr(T a) { ostringstream oss_; oss_ << a; return oss_.str(); }
inline int toInt(string s) { return atoi(s.c_str()); }

// prime
bool isPrime(int a) { for(int i=2; i*i <=a; i++) if(a%i == 0) return false; return true; }

int const dx[] = {-1,0,1,0,-1,1,1,-1};
int const dy[] = {0,-1,0,1,-1,-1,1,1};

//////////////////////////////////////////////////////////////


int FairField(int y,int m,int d)
{
  // adjust
  if(m <= 2) {
    y --;
    m += 12;
  }
  
  int U = y/4 - y/100 + y/400;
  return 365*y + U + 153*(m+1)/5 + d - 428;
}

int main() {
  int y1, m1, d1, y2, m2, d2;
  
  while(cin >> y1 >> m1 >> d1 >> y2 >> m2 >> d2) {
    if(y1 == -1) break;
    
    cout << FairField(y2,m2,d2) - FairField(y1,m1,d1) << endl;
  }
  
  return 0;
}