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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>

using namespace std;

// repetition
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

// container util
#define all(x) (x).begin(), (x).end()

// debug
#define dump(x) cerr << #x << " = " << (x) << endl;
#define debug(x)                                         \
  cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" \
       << " " << __FILE__ << endl;

// typedef
typedef long long lint;
typedef unsigned long long ull;
typedef complex<long double> Complex;
typedef pair<int, int> P;
typedef tuple<int, int, int> TP;
typedef vector<int> vec;
typedef vector<vec> mat;

// constant
const int MOD = (int)1e9 + 7;
const int INF = (int)1e18;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int ddx[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[] = {1, 1, 0, -1, -1, -1, 0, 1};

// conversion
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}

//
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
//

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int N;
  cin >> N;

  for (int i = 1; i < 10; i++) {
    for (int j = 1; j < 10; j++) {
      if (i * j == N) {
        cout << "Yes\n";
        return 0;
      }
    }
  }
  cout << "No\n";

  return 0;
}