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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("Ofast")

//#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,fma")
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstdlib>
#include<set>
#include<vector>
#include<sstream>
#include<queue>
#include <iomanip>
#include<unordered_set>
#include <stack>
#include <unordered_map>
#include <map>
constexpr auto INF = 9223372036854775807;
typedef long long int ll;
typedef unsigned long long int ull;
typedef unsigned long int ul;
#define f(i,a,b) for(ll i=(ll)a;i<(ll)b;i+=1)
#define rf(i,a,b) for(ll i=(ll)a;i>=(ll)b;i-=1)
#define endl '\n'
#define N 1000000007 // prime modulo value 
#define M 998244353
#define all(x) x.begin(),x.end()
using namespace std;
// From Geeksforgeeks
inline ll gcd(ll a, ll b)
{
	if (b == 0)
		return a;
	return gcd(b, a % b);
}
// Power function took from geeksforgeeks
ll power(ll x, ll y)
{
	ll res = 1;

	x = x%N;
	if (x == 0) return 0;
	while (y > 0)
	{
		if (y & 1)
			res = (res * x)%N;
		y = y >> 1;
		x = (x * x)%N;
	}
	return res;
}
double dist(ll x1,ll y1, ll x2,ll y2)
{
	double ans = sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
	return ans;
}
int main()
{
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	ll n;
	cin >> n;
	f(i, 9, 1)
	{
		if (n % i == 0)
		{
			n /= i;
			break;
		}
	}
	if (n < 10)
	{
		cout << "Yes";
	}
	else
		cout << "No";
	return 0;
}