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<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<cstdlib>
#include<iomanip>
#include<queue>
#include<set>

using namespace std;

bool isLeap(int year)
{
    if (year % 400 == 0)
    {
        return true;
    }
    
    if (year % 100 == 0)
    {
        return false;
    }
    
    return (year % 4 == 0);
}

long long dateSerial(int year, int month, int day)
{
    static int days[12] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
    
    long long res = day;
    
    for (int i = 0; i < month - 1; i++)
    {
        res += days[i];
        
        if(isLeap(year) && i == 1)
        {
            res++;
        }
    }
    
    for (int i = 1; i < year; i++)
    {
        res += 365;
       
        if (isLeap(i))
        {
            res++;
        }
    }
    
    return res;
}

void Solution()
{
    int y1, m1, d1, y2, m2, d2;
    
    while (cin>>y1>>m1>>d1>>y2>>m2>>d2)
    {
        if (y1 < 0 || m1 < 0 || d1 < 0 || y2 < 0 || m2 < 0 || d2 < 0)
        {
            break;
        }
        
        long long date1 = dateSerial(y1, m1, d1);
        long long date2 = dateSerial(y2, m2, d2);
        
        cout<<date2 - date1<<endl;
    }
}

int main()
{
    Solution();
    return 0;
}