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>
#include<cstdio>
#include <cstring>
#include<map>
#include<string>
#include<stack>
#include <algorithm>
#include <strstream>
#include<vector>
#include<map>
#include<cmath>
using namespace std;

int GetDays(int y, int m, int d)
{
if (m <= 2)y-- , m += 12;
int dy = 365 * (y - 1);
int c = y / 100;
int dl = (y >> 2) - c + (c >> 2);
int dm = (m * 979 - 1033) >> 5;
return dy + dl + dm + d - 1;
}

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