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: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef unsigned uint;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldbl;
typedef pair<int, int> pii;
typedef pair<uint, uint> puu;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pull;
typedef pair<double, double> pdd;
typedef vector<int> vi;
typedef vector<uint> vu;
typedef vector<ll> vll;
typedef vector<ull> vull;
typedef vector<pii> vpii;
typedef vector<puu> vpuu;
typedef vector<pll> vpll;
typedef vector<pull> vpull;
typedef vector<string> vstr;
typedef vector<double> vdbl;
typedef vector<ldbl> vldbl;
#define pb push_back
#define ppb pop_back
#define pfr push_front
#define ppfr pop_front
#define emp emplace
#define empb emplace_back
#define be begin
#define rbe rbegin
#define all(x) (x).be(), (x).end()
#define rall(x) (x).rbe(), (x).rend()
#define fir first
#define sec second
#define mkp make_pair
#define brif(cond) if (cond) break
#define ctif(cond) if (cond) continue
#define retif(cond) if (cond) return
void canhazfast() {ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);}
template<typename T> T gcd(T a, T b) {return b ? gcd(b, a%b) : a;}
template<typename T> T extgcd(T a, T b, T &x, T &y)
{
    T x0 = 1, y0 = 0, x1 = 0, y1 = 1;
    while (b) {
        T q = a/b; a %= b; swap(a, b);
        x0 -= q*x1; swap(x0, x1);
        y0 -= q*y1; swap(y0, y1);
    }
    x = x0; y = y0; return a;
}
int ctz(uint x) {return __builtin_ctz(x);}
int ctzll(ull x) {return __builtin_ctzll(x);}
int clz(uint x) {return __builtin_clz(x);}
int clzll(ull x) {return __builtin_clzll(x);}
int popcnt(uint x) {return __builtin_popcount(x);}
int popcntll(ull x) {return __builtin_popcountll(x);}
int bsr(uint x) {return 31^clz(x);}
int bsrll(ull x) {return 63^clzll(x);}

int main()
{
    canhazfast();

    int n;

    cin >> n;
    for (int i = 1; i < 10; ++i) {
        retif(n%i == 0 && n/i < 10) cout << "Yes\n", 0;
    }
    cout << "No\n";

    return 0;
}