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
//10
#include<iostream>

using namespace std;

int dc(int y,int m,int d){
  int cy,cm,cd;
  cy=cm=cd=1;
  int dct=0;
  while(y!=cy||m!=cm||cd!=d){
    dct++;
    cd++;
    static const int dom[][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31},{0,31,29,31,30,31,30,31,31,30,31,30,31}};
    if(cd>dom[cy%400==0||(cy%4==0&&cy%100)][cm]){
      cd=1;
      cm++;
      if(cm>12){
	cm=1;
	cy++;
      }
    }
  }
  return dct;
}

int main(){
  for(int y,m,d,yy,mm,dd;cin>>y>>m>>d>>yy>>mm>>dd,(y|m|d|yy|mm|dd)>=0;){
    cout<<dc(yy,mm,dd)-dc(y,m,d)<<endl;
  }
  return 0;
}