B. Nirvana Codeforces Round #549 (Div. 2) （递归dfs）

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Kurt reaches nirvana when he finds the product of all the digits of some positive integer. Greater value of the product makes the nirvana deeper.

Help Kurt find the maximum possible product of digits among all integers from 1 to n.

Input

The only input line contains the integer nn (1≤n≤2⋅10⁹).

Output

Print the maximum product of digits among all integers from 1 to n.

Examples

Input

Copy

390

Output

Copy

216

Input

Copy

7

Output

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7

Input

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1000000000

Output

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387420489

Note

In the first example the maximum product is achieved for 389389 (the product of digits is 3⋅8⋅9=216).

In the second example the maximum product is achieved for 77 (the product of digits is 7).

In the third example the maximum product is achieved for 99999999 (the product of digits is 9⁹=38742048999).

题意：给出n，找出不大于n的一个数，试其乘积最大。

思路：对于一个位置上的数，①可以保持不变，②可以使其变成9，前置位-1.

这样我们可以递归枚举。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3
 4 int n;
 5
 6 int cal(int n)
 7 {
 8     if(n == 0)return 1;
 9     else if(n < 10)return n;
10     else
11     {
12         return max(cal(n/10)*(n%10),cal(n/10-1)*9);
13     }
14 }
15
16 int main()
17 {
18     scanf("%d",&n);
19     printf("%d\n",cal(n));
20 }

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原文地址：https://www.cnblogs.com/iwannabe/p/10687190.html