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 <iostream>
using namespace std;
int getday(int, int, int);

int main(void)
{
  int y1, m1, d1, y2, m2, d2;
  int day1, day2;
  int year;
  
  while(1) {
    cin >> y1 >> m1 >> d1 >> y2 >> m2 >> d2;
    if(y1 < 0 || m1 < 0 || d1 < 0 || y2 < 0 || m2 < 0 || d2 < 0) break;
    day1 = getday(y1, m1, d1);
    day2 = getday(y2, m2, d2);
    
    cout << day2 - day1 << endl;
  }
  return 0;
}

int getday(int y, int m, int d)
{
  int day = 0,u;

  u = (y % 4 == 0 && y % 100 != 0 || y % 400 == 0);

  for(int i = 1 ; i < y ; i++) {
    day += 365;
    if(i % 4 == 0 && i % 100 != 0 || i % 400 == 0) day += 1;
      }

  for(int i = 1 ; i < m ; i++) {
    day += 30;
    if(i == 1 || i == 3 || i == 5 || i == 7 || i == 8 || i == 10 || i == 12) day++;
    if(i == 2) { day -= 2; if(u) day++; }
  }
  for(int i = 0 ; i < d ; i++) {
    day++;
  }
  return day;
}