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;
long long mul(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  return (a % p + b % p) % p;
}
void input(long long a[], long long sz) {
  for (long long i = 0; i < sz; i++) cin >> a[i];
}
void print(long long a[], long long sz) {
  for (long long i = 0; i < sz; i++) {
    if (i == sz - 1)
      cout << a[i] << "\n";
    else
      cout << a[i] << " ";
  }
}
long long maxr(long long a[], long long sz) {
  long long ma;
  for (long long i = 0; i < sz; i++) {
    if (i == 0)
      ma = a[i];
    else if (a[i] > ma)
      ma = a[i];
  }
  return ma;
}
long long minr(long long a[], long long sz) {
  long long mi;
  for (long long i = 0; i < sz; i++) {
    if (i == 0)
      mi = a[i];
    else if (a[i] < mi)
      mi = a[i];
  }
  return mi;
}
long long isprm(long long n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  return 1;
}
long long power(long long x, long long y, long long p = 1000000007) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, long long p = 1000000007) {
  return power(n, p - 2, p);
}
long long ncrMod(long long n, long long r, long long p = 1000000007) {
  if (r == 0) return 1;
  long long fac[n + 1];
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
struct comp {
  bool operator()(const pair<long long, long long> &a,
                  const pair<long long, long long> &b) const {
    if (a.first == b.first) {
      return (a.second > b.second);
    }
    return (a.first < b.first);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  string s;
  cin >> s;
  long long x = s.length();
  long long a[x + 5];
  a[0] = (long long)s[0] - '0';
  for (long long i = 1; i < x; i++) {
    long long d = (long long)s[i] - '0';
    a[i] = a[i - 1] * d;
  }
  long long ans = a[x - 1];
  long long k = 9;
  for (long long i = x - 1; i >= 1; i--) {
    long long d = (long long)s[i - 1] - '0';
    d--;
    if (d == 0 && i == 1) d = 1;
    if (i == 1)
      ans = max(ans, (d * k));
    else
      ans = max(ans, (a[i - 2] * d * k));
    k *= 9;
  }
  cout << ans << "\n";
  return 0;
}
