Seven Segment Display（数位dp）

A seven segment display, or seven segment indicator, is a form of electronic display device for displaying decimal numerals that is an alternative to the more complex dot matrix displays. Seven segment displays are widely used in digital clocks, electronic meters, basic calculators, and other electronic devices that display numerical information.

Edward, a student in Marjar University, is studying the course "Logic and Computer Design Fundamentals" this semester. He bought an eight-digit seven segment display component to make a hexadecimal counter for his course project.

In order to display a hexadecimal number, the seven segment display component needs to consume some electrical energy. The total energy cost for display a hexadecimal number on the component is the sum of the energy cost for displaying each digit of the number. Edward found the following table on the Internet, which describes the energy cost for display each kind of digit.

For example, in order to display the hexadecimal number "5A8BEF67" on the component for one second, 5 + 6 + 7 + 5 + 5 + 4 + 6 + 3 = 41 units of energy will be consumed.

Edward's hexadecimal counter works as follows:

• The counter will only work for n seconds. After n
• At the beginning of the 1st second, the counter will begin to display a previously configured eight-digit hexadecimal number m.
• At the end of the i-th second (1 ≤ i < n), the number displayed will be increased by 1. If the number displayed will be larger than the hexadecimal number "FFFFFFFF" after increasing, the counter will set the number to 0 and continue displaying.

Given n and m, Edward is interested in the total units of energy consumed by the seven segment display component. Can you help him by working out this problem?

Input

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 10⁵), indicating the number of test cases. For each test case:

The first and only line contains an integer n (1 ≤ n ≤ 10⁹) and a capitalized eight-digit hexadecimal number m (00000000 ≤ m

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 output one line, indicating the total units of energy consumed by the eight-digit seven segment display component.

Sample Input

3 5 89ABCDEF 3 FFFFFFFF 7 00000000

Sample Output

208 124 327

Hint

For the first test case, the counter will display 5 hexadecimal numbers (89ABCDEF, 89ABCDF0, 89ABCDF1, 89ABCDF2, 89ABCDF3) in 5 seconds. The total units of energy cost is (7 + 6 + 6 + 5 + 4 + 5 + 5 + 4) + (7 + 6 + 6 + 5 + 4 + 5 + 4 + 6) + (7 + 6 + 6 + 5 + 4 + 5 + 4 + 2) + (7 + 6 + 6 + 5 + 4 + 5 + 4 + 5) + (7 + 6 + 6 + 5 + 4 + 5 + 4 + 5) = 208.

For the second test case, the counter will display 3 hexadecimal numbers (FFFFFFFF, 00000000, 00000001) in 3 seconds. The total units of energy cost is (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4) + (6 + 6 + 6 + 6 + 6 + 6 + 6 + 6) + (6 + 6 + 6 + 6 + 6 + 6 + 6 + 2) = 124.

题目大概：

每个十六进制的一位数都有一个价值，算出一个八位的十六进制的价值，并算出随后的n-1个十六进制的数的价值。计算总价值。到达最大后。从零开始。

思路：

刚开始用的模拟，写了一半想起了数位dp，但是不甘心的写完了模拟。确实TLE了。

用数位dp，当然也是有过TLE，只需把dp数组的更新拿到外面就可以了，因为dp数组的值不需要更新，每个数计算的同一个dp数组的值是一样的。

主要是16进制，搞了好久，具体的数位dp过程是最简单的数位dp，还有可能TLE要注意。

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define LL long long
int a[20];
long long dp[22][2222];
int b[16]={6,2,5,5,4,5,6,3,7,6,6,5,4,5,5,4};
long long sove(int pos,int sum,int limit)
{
    if(pos<0)return sum;
    if(!limit&&dp[pos][sum]!=-1)return dp[pos][sum];
    int end=limit?a[pos]:15;
    LL su=0;
    for(int i=0;i<=end;i++)
    {
        su+=sove(pos-1,sum+b[i],limit&&(i==a[pos]));
    }
    if(!limit)dp[pos][sum]=su;
    return su;
}

long long go(long long x)
{

    for(int i=0;i<8;i++)
    {
        a[i]=x%16;
        x/=16;
    }
    return sove(7,0,1);
}
int main()
{
    long long n,m,l;
    int t;
    //4294967295

    scanf("%d",&t);
    memset(dp,-1,sizeof(dp));
    while(t--)
    {

        scanf("%lld %llX",&n,&m);
        n--;
        l=m+n;
        if(l>(LL)4294967295)
        {
            l%=(LL)(4294967296);
            printf("%lld\n",go((LL)4294967295)+go(l)-go(m-1));
        }
        else
        {
            printf("%lld\n",go(l)-go(m-1));
        }


    }
    return 0;
}