How Many Nines

If we represent a date in the format YYYY-MM-DD (for example, 2017-04-09), do you know how many 9s will appear in all the dates between Y₁-M₁-D₁ and Y₂-M₂-D₂ (both inclusive)?

Note that you should take leap years into consideration. A leap year is a year which can be divided by 400 or can be divided by 4 but can't be divided by 100.

Input

The first line of the input is an integer T (1 ≤ T ≤ 10⁵), indicating the number of test cases. Then T test cases follow. For each test case:

The first and only line contains six integers Y₁, M₁, D₁, Y₂, M₂, D₂, their meanings are described above.

It's guaranteed that Y₁-M₁-D₁ is not larger than Y₂-M₂-D₂. Both Y₁-M₁-D₁ and Y₂-M₂-D₂are between 2000-01-01 and 9999-12-31, and both dates are valid.

We kindly remind you that this problem contains large I/O file, so it's recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++.

<h4< dd="">Output

For each test case, you should output one line containing one integer, indicating the answer of this test case.

<h4< dd="">Sample Input

4
2017 04 09 2017 05 09
2100 02 01 2100 03 01
9996 02 01 9996 03 01
2000 01 01 9999 12 31

<h4< dd="">Sample Output

4
2
93
1763534

Hint

For the first test case, four 9s appear in all the dates between 2017-04-09 and 2017-05-09. They are: 2017-04-09 (one 9), 2017-04-19 (one 9), 2017-04-29 (one 9), and 2017-05-09 (one 9).

For the second test case, as year 2100 is not a leap year, only two 9s appear in all the dates between 2100-02-01 and 2100-03-01. They are: 2017-02-09 (one 9) and 2017-02-19 (one 9).

For the third test case, at least three 9s appear in each date between 9996-02-01 and 9996-03-01. Also, there are three additional nines, namely 9996-02-09 (one 9), 9996-02-19 (one 9) and 9996-02-29 (one 9). So the answer is 3 × 30 + 3 = 93.

      Question: Given two dates, find the number of nines between the two dates;

      Idea: Let a[i][j][k] be the number of 9s from January 01, 2000 to i year j month k day, first calculate all the conditions, and finally subtract the two dates. .

      Code:

      

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<cstring>
#include<iostream>
#include<set>
#include<queue>
#include<functional>
#include<string>
#include<map>

using namespace std;
int judge(int x)//judgment how many 9s
{
	int n;
	int flag = 0;
	while (x > 0)
	{
		n = x % 10;
		if (n == 9)
			flag++;
		x = x / 10;
	}
	return flag;
}
bool isleap(int x)
{
	if ((x % 4 == 0 && x % 100 != 0) || x % 400 == 0)
		return true;
	else
		return false;
}
int days[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

int leapdays[13] = {0,31,29,31,30,31,30,31,31,30,31,30,31};
int a[10000][13][32];
intmain()
{
	int x;
		int t;
		scanf("%d", &t);
		int i, j, k;
		int years;
		int y1, m1, d1, y2, m2, d2;
		a[2000][1][1] = 0;
		int pre = 0;
		for (i = 2000; i <= 9999; i++)
		{
			if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0)
			{
				for (j = 1; j <= 12; j++)
				{
					for (k = 1; k <= leapdays[j]; k++)
					{
						 x = i * 10000 + j * 100 + k;//Determine how many 9s there are on this day
						a[i][j][k] = judge(x) + pre;
						pre = a [i] [j] [k];
					}
				}
			}
			else
			{
				for (j = 1; j <= 12; j++)
				{
					for (k = 1; k <= days[j]; k++)
					{
                        x = i * 10000 + j * 100 + k;
						a[i][j][k] = judge(x) + pre;
						pre = a [i] [j] [k];
					}

				}
			}
		}

	for (int i = 0; i < t; i++)
	{
		scanf("%d%d%d", &y1, &m1, &d1);
		scanf("%d%d%d", &y2, &m2, &d2);
		int x = y1 * 10000 + m1 * 100 + d1;
		ans = judge(x);
		printf("%d\n", a[y2][m2][d2] - a[y1][m1][d1]+ans);
	}
	return 0;
}