Problem:
Score : 200 points

Problem Statement

Having learned the multiplication table, Takahashi can multiply two integers between 1 and 9 (inclusive) together.

Given an integer N, determine whether N can be represented as the product of two integers between 1 and 9. If it can, print Yes; if it cannot, print No.

Constraints

1 \leq N \leq 100

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If N can be represented as the product of two integers between 1 and 9 (inclusive), print Yes; if it cannot, print No.

Sample Input 1

10

Sample Output 1

Yes

10 can be represented as, for example, 2 \times 5.

Sample Input 2

50

Sample Output 2

No

50 cannot be represented as the product of two integers between 1 and 9.

Sample Input 3

81

Sample Output 3

Yes

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

/* ex3_4 KitamoriFumiya */
#include <stdio.h>

int main(void){
	
	int dividend, divisor, quotient, surplus; // dividend(割られる数), divisor(割る数), quotient(商), surplus(余り)をint型の変数として宣言
	scanf("%d", &dividend); // dividendを入力
	
	divisor = 1; // 割る数の初期値を1にする
	// dividendを1から9までの整数で割る
	while (divisor < 10){
		quotient = dividend / divisor; // 商を求める
		surplus = dividend % divisor; // 余りを求める
		// もし1から9まで順にdividendを割っていた時にdivindendが割り切れ、かつ商が1以上9以下なら途中でループを抜け出す
		if ( 0 < quotient && quotient < 10 && surplus == 0) break;
		divisor++; // 割る数を1ずつ増やしていく
	}
	
	// もし1から9まで順にdividendを割っていた時にdivindendが割り切れ、かつ商が1以上9以下ならdividendを1以上9以下の2つの整数の積で表せるのでYesを出力
	if ( 0 < quotient && quotient < 10 && surplus == 0){
		printf("Yes\n");
	}
	// それ以外はNoを出力
	else{
		printf("No\n");
	}
	
	return 0;
}