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<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<n;i++)
using namespace std;

bool is_uruu(int y) {
	bool uruu = false;
	if (!(y % 4))uruu = true;
	if (!(y % 100))uruu = false;
	if (!(y % 400))uruu = true;
	return uruu;
}
int main() {
	int y1, m1, d1, y2, m2, d2;
	int m[]{ 31,28,31,30,31,30,31,31,30,31,30,31 }, uruu_m[]{ 31,29,31,30,31,30,31,31,30,31,30,31 };
	while (cin >> y1 >> m1 >> d1 >> y2 >> m2 >> d2, y1 >= 0 && m1 >= 0 && d1 >= 0 && y2 >= 0 && m2 >= 0 && d2 >= 0) {
		int cnt = 0;
		for (int i = y1; i < y2; i++) {
			cnt += 365;
			if (is_uruu(i))cnt++;
		}
		if (is_uruu(y1)) {
			for (int i = 0; i < m1 - 1; i++)cnt -= uruu_m[i];
			cnt -= d1 - 1;
		}
		else {
			for (int i = 0; i < m1 - 1; i++)cnt -= m[i];
			cnt -= d1 - 1;
		}
		if (is_uruu(y2)) {
			for (int i = 0; i < m2 - 1; i++)cnt += uruu_m[i];
			cnt += d2;
		}
		else {
			for (int i = 0; i < m2 - 1; i++)cnt += m[i];
			cnt += d2;
		}
		cout << cnt - 1 << endl;
	}
}