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<stdio.h>
int main(void){
	int y1,m1,d1,y2,m2,d2;
	int i;
	int y1day,y2day;
	while(1){
		scanf("%d %d %d %d %d %d",&y1,&m1,&d1,&y2,&m2,&d2);
		y1day=y2day=0;
		if(y1==-1 || m1==-1 || d1==-1 || y2==-1 || m2==-1  || d2==-1) break;
		y1day=y2day=0;
		for(i=1;i<y1;i++){
			y1day+=365;
			if((i%4==0 && i%100!=0) || i%400==0) y1day++;
		}
		for(i=1;i<m1;i++){
			y1day+=31;
			if(i==4 || i==6 || i==9 || i==11) y1day-=1;
			if(i==2){
				y1day-=3;
				if((y1%4==0 && y1%100!=0) || y1%400==0) y1day++;
			}
		}
		y1day+=d1;
		for(i=1;i<y2;i++){
			y2day+=365;
			if((i%4==0 && i%100!=0) || i%400==0) y2day++;
		}
		for(i=1;i<m2;i++)	{
			y2day+=31;
			if(i==4 || i==6 || i==9 || i==11) y2day--;
			if(i==2){
				y2day-=3;
				if((y2%4==0 && y2%100!=0) || y2%400==0) y2day++;
			}
	//		printf("y2d==%d\n",y2day);
		}
		y2day+=d2;
	//	printf("%d %d\n",y1day,y2day);
		printf("%d\n",y2day-y1day);
	}
	return 0;
}