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 <sstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <climits>
#include <cfloat>
using namespace std;

int main()
{
    int month[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

    for(;;){
        int y[2], m[2], d[2];
        for(int i=0; i<2; ++i){
            cin >> y[i] >> m[i] >> d[i];
            if(y[i] < 0 || m[i] < 0 || d[i] < 0)
                return 0;
        }

        long long sum[2] = {0};
        for(int i=0; i<2; ++i){
            for(int j=1; j<y[i]; ++j){
                sum[i] += 365;
                if(j % 400 == 0 || (j % 100 != 0 && j % 4 == 0))
                    ++ sum[i];
            }
            for(int j=1; j<m[i]; ++j){
                sum[i] += month[j];
                if(j == 2 && (y[i] % 400 == 0 || (y[i] % 100 != 0 && y[i] % 4 == 0)))
                    ++ sum[i];
            }
            sum[i] += d[i];
        }

        cout << (sum[1] - sum[0]) << endl;
    }
}