A. Sequence with Digits
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Let’s define the following recurrence:
an+1=an+minDigit(an)⋅maxDigit(an).
Here minDigit(x) and maxDigit(x) are the minimal and maximal digits in the decimal representation of x without leading zeroes. For examples refer to notes.

Your task is calculate aK for given a1 and K.

Input
The first line contains one integer t (1≤t≤1000) — the number of independent test cases.

Each test case consists of a single line containing two integers a1 and K (1≤a1≤1018, 1≤K≤1016) separated by a space.

Output
For each test case print one integer aK on a separate line.

Example
inputCopy
8
1 4
487 1
487 2
487 3
487 4
487 5
487 6
487 7
outputCopy
42
487
519
528
544
564
588
628
Note
a1=487

a2=a1+minDigit(a1)⋅maxDigit(a1)=487+min(4,8,7)⋅max(4,8,7)=487+4⋅8=519

a3=a2+minDigit(a2)⋅maxDigit(a2)=519+min(5,1,9)⋅max(5,1,9)=519+1⋅9=528

a4=a3+minDigit(a3)⋅maxDigit(a3)=528+min(5,2,8)⋅max(5,2,8)=528+2⋅8=544

a5=a4+minDigit(a4)⋅maxDigit(a4)=544+min(5,4,4)⋅max(5,4,4)=544+4⋅5=564

a6=a5+minDigit(a5)⋅maxDigit(a5)=564+min(5,6,4)⋅max(5,6,4)=564+4⋅6=588

a7=a6+minDigit(a6)⋅maxDigit(a6)=588+min(5,8,8)⋅max(5,8,8)=588+5⋅8=628

给定一个数 N N N,定义操作 o p op op
op:
N = N + m i n ( N 的 每 一 位 数 字 ) ∗ m a x ( N 的 每 一 位 数 字 ) N=N+min(N的每一位数字)*max(N的每一位数字) N=N+min(N的每一位数字)∗max(N的每一位数字)
可以发现,多次操作后 N N N可能出现某一位出现 0 0 0,所以,min(N的每一位数字)*max(N的每一位数字)为0,即:出现0后无论怎么操作 N N N都不会改变
所以,执行 K K K次模拟,当 N N N出现0的时候break掉

//#define debug
#ifdef debug
#include <time.h>
#include "/home/majiao/mb.h"
#endif
 
 
#include <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <math.h>
#define MAXN ((int)1e5+7)
#define ll long long int
#define INF (0x7f7f7f7f)
#define int long long int
#define QAQ (0)
 
using namespace std;
 
#ifdef debug
#define show(x...) \ do { \ cout << "\033[31;1m " << #x << " -> "; \ err(x); \ } while (0)
void err() { cout << "\033[39;0m" << endl; }
#endif
 
template<typename T, typename... A>
void err(T a, A... x) { cout << a << ' '; err(x...); }
 
int n, m, Q, K;
 
int minD(int x) {
	int ret = 99;
	while(x) {
		ret = min(ret, x%10);
		x /= 10;
	}
	return ret;
}
 
int maxD(int x) {
	int ret = 0;
	while(x) {
		ret = max(ret, x%10);
		x /= 10;
	}
	return ret;
}
 
int calc() {
	while(--K) {
		int tmin = minD(n), tmax = maxD(n);
// show(n, tmin, tmax);
		if(!tmin) break;
		n = n + (tmin * tmax);
	}
	return n;
}
 
signed main() {
#ifdef debug
	freopen("test", "r", stdin);
	clock_t stime = clock();
#endif
	scanf("%lld ", &Q);
	while(Q--) {
		scanf("%lld %lld ", &n, &K);
		int ans = calc();
		printf("%lld\n", ans);
	}
 
 
 
#ifdef debug
	clock_t etime = clock();
	printf("rum time: %lf 秒\n",(double) (etime-stime)/CLOCKS_PER_SEC);
#endif 
	return 0;
}