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<string>
#include<vector>
using namespace std;

int main() 
{
	while (1) {
		int a, b, c, d, e, f;
		cin >> a >> b >> c >> d >> e >> f;
		if (a < 0 || b < 0 || c < 0 || d < 0 || e < 0 || f < 0)break;
		int l, r; l = r = 0;
		for (int g = 1; g < a; g++) {
			if (g % 400 == 0) l += 366;
            else	if (g % 100 == 0)l += 365;
            else    if (g % 4 == 0) l += 366;
            else l += 365;
		}
		for (int g = 1; g < b; g++) {
			switch (g) {
			case 1:case 3:case 5:case 7:case 8:case 10:case 12:l += 31; break;
			case 4:case 6:case 9:case 11: l += 30; break;
			case 2:if (a % 400 == 0) l += 29;
				   else	if (a % 100 == 0)l += 28;
				   else    if (a % 4 == 0) l += 29;
				   else l += 28;
			}
		}
		l += c;
		for (int g = 1; g < d; g++) {
			if (g % 400 == 0) r += 366;
			else	if (g % 100 == 0)r += 365;
			else    if (g % 4 == 0) r += 366;
			else r += 365;
		}
		for (int g = 1; g < e; g++) {
			switch (g) {
			case 1:case 3:case 5:case 7:case 8:case 10:case 12:r += 31; break;
			case 4:case 6:case 9:case 11: r += 30; break;
			case 2:if (d % 400 == 0) r += 29;
				   else	if (d% 100 == 0)r += 28;
				   else    if (d % 4 == 0) r+= 29;
				   else r += 28;
			}
		}
		r += f;
		cout << r - l << endl;
	}
}