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 betweenY₁-M₁-D₁ andY₂-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. ThenT 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 thanY

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++.

Output

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

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

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.

做這種題最好就是f(b)-f(a-1)這樣子算一次。然後打表。

這道題我卡了兩次，不知道是掛在輸入上還是掛在cout上，於是我果斷寫了讀入掛，就AC了。

#include <stdio.h>
#include <string.h>
#include <ctime>
#include <stack>
#include <string>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <map>
#include <ctime>
#include <queue>
#include <vector>
using namespace std;
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lowbit(x) (x&-x)

const int maxn = 1e6+5;
const int INF = 0x3f3f3f3f;

int s[366*8005];
int yd[10005];

bool isRun(int year)
{
    if(year%100==0)
    {
        if(year%400==0)
            return true;
    }
    else if(year%4==0)
        return true;
    return false;
}

void del(int& year,int& month,int& day)
{
    int a[] ={31,31,28,31,30,31,30,31,31,30,31,30,31};
    if(isRun(year)) a[2]++;
    day--;
    if(day==0)
    {
        day = a[month-1];
        month--;
        if(month==0)
        {
            month = 12;
            year--;
        }
    }
}

int getNum(int year)
{
    int ret = 0;
    while(year)
    {
        if(year%10==9) ret++;
        year/=10;
    }
    return ret;
}

void init()
{
    for(int i=2000; i<=9999; i++)
        if(isRun(i))
            yd[i] = yd[i-1]+getNum(i)*366+36+30;
        else
            yd[i] = yd[i-1]+getNum(i)*365+35+30;
}

int f(int year,int month,int day)
{
    int a[] ={31,31,28,31,30,31,30,31,31,30,31,30,31};
    if(isRun(year)) a[2]++;
    if(year<2000) return 0;

    int ret = yd[year-1];
    ret += (month-1)*3;
    if(month>2&&!isRun(year)) ret--;
    if(day>=9&&day<19)
        ret++;
    else if(day>=19&&day<29)
        ret+=2;
    else if(day>=29)
        ret+=3;

    int dat = 0;
    for(int i=1; i<month; i++)
        dat += a[i];
    dat+=day;
    ret += getNum(year)*dat;
    if(month>9)
        ret += 30;
    else if(month==9)
        ret += day;
    return ret;
}

int readint()
{
    int ret = 0;
    char c = getchar();
    while (c<'0' || c>'9') c = getchar();
    while (c >= '0'&&c <= '9')
    {
        ret = ret * 10 + c - '0';
        c = getchar();
    }
    return ret;
}

int main()
{
    int T;
    init();
    scanf("%d",&T);
    while(T--)
    {
        int y1,y2,m1,m2,d1,d2;
        y1 = readint();
        m1 = readint();
        d1 = readint();
        y2 = readint();
        m2 = readint();
        d2 = readint();

        del(y1,m1,d1);
        printf("%d\n",f(y2,m2,d2)-f(y1,m1,d1));
    }
    return 0;
}