Kurt reaches nirvana when he finds the product of all the digits of some positive integer. Greater value of the product makes the nirvana deeper.

Help Kurt find the maximum possible product of digits among all integers from 1 to n.

Input

The only input line contains the integer n (1 ≤ n ≤ 2⋅10^9).

Output

Print the maximum product of digits among all integers from 1 to n.

Examples

Input


390


Output


216


Input


7


Output


7


Input


1000000000


Output


387420489

Note

In the first example the maximum product is achieved for 389 (the product of digits is 3⋅8⋅9=216).

In the second example the maximum product is achieved for 7 (the product of digits is 7).

In the third example the maximum product is achieved for 999999999 (the product of digits is 9^9=387420489).
#include <bits/stdc++.h>
using namespace std;
default_random_engine generator;
uniform_int_distribution<int> distribution(1, INT_MAX);
vector<vector<long long> > one(int sz) {
  vector<vector<long long> > ret;
  ret.resize(sz);
  for (int i = 0; i < ret.size(); i++) {
    ret[i].resize(sz);
    for (int j = 0; j < ret.size(); j++) {
      ret[i][j] = (i == j);
    }
  }
  return ret;
}
vector<vector<long long> > mult(vector<vector<long long> > a,
                                vector<vector<long long> > b) {
  vector<vector<long long> > ret;
  ret.resize(a.size());
  for (int i = 0; i < ret.size(); i++) {
    ret[i].resize(a.size());
    for (int j = 0; j < ret.size(); j++) {
      ret[i][j] = 0;
    }
  }
  for (int k = 0; k < ret.size(); k++) {
    for (int i = 0; i < a.size(); i++) {
      for (int j = 0; j < b.size(); j++) {
        ret[i][j] = (ret[i][j] + a[i][k] * b[k][j]);
        ret[i][j] %= 998244353;
      }
    }
  }
  return ret;
}
vector<vector<long long> > pw(vector<vector<long long> > a, long long b) {
  if (b == 0) return one((int)a.size());
  if (b % 2) return mult(a, pw(a, b - 1));
  return pw(mult(a, a), b / 2);
}
long long my_rand(long long a = 0, long long b = INT_MAX) {
  return a + distribution(generator) % b;
}
inline int mul(int a, int b) { return int(a * 1ll * b % 998244353); }
inline int norm(int a) {
  if (a >= 998244353) a -= 998244353;
  if (a < 0) a += 998244353;
  return a;
}
inline int binPow(int a, int k) {
  int ans = 1;
  while (k > 0) {
    if (k & 1) ans = mul(ans, a);
    a = mul(a, a);
    k >>= 1;
  }
  return ans;
}
inline int subtract(int a, int b) { return norm(a - b); }
inline int sum(int a, int b) { return int((a + b) % 998244353); }
inline int inv(int a) { return binPow(a, 998244353 - 2); }
long long ceil_int(long long a, long long b) { return a / b + (a % b != 0); }
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int main() {
  generator.seed(chrono::system_clock::now().time_since_epoch().count());
  cin.sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  string t;
  int len = s.length();
  long long res = 1;
  long long cur = 1;
  for (int i = 0; i < (1 << len); i++) {
    t = s;
    cur = 1;
    for (int j = len - 1; j >= 0; j--) {
      if (i & (1 << j) && j > 0) {
        cur *= 9ll;
        t[j - 1] = t[j - 1] - 1;
      } else if (j != 0 || t[j] != '0') {
        cur *= (long long)(t[j] - '0');
      }
    }
    res = max(res, cur);
  }
  cout << res;
  return 0;
}
