n=int(input())
primes=[2,3,5,7,11,13,17,19,23,29]
def f(n,i):
if n==1:
return 1
if i==10:
return 10**18
ans=10**18
for j in range(2,n+1):
if n%j==0:
ans=min(ans,primes[i]**(j-1)*f(n//j,i+1))
return ans
print(f(n,0))#include<bits/stdc++.h>
#include<iostream>
using namespace std;
typedef long long ll;
typedef long double ld;
const int N = 2e3 + 5;
const long long INF = 1e18 + 1;
int n,m,q,timer,k;
long long dp[N + 1][N + 1];
long long a[N + 1][N + 1];
int pw[N + 1];
int p[9] = {2,3,5,7,11,13,17,19,23};
long long f(int n,int k){
if(n == 1) return 1;
if(k < 0) return INF;
if(dp[n][k]) return dp[n][k];
long long ans = f(n,k - 1);
for(int i = 0; i <= pw[p[k]]; i++){
if(n % (i + 1) == 0){
long long to = f(n / (i + 1) , k - 1);
if(to < INF / a[p[k]][i]){
ans = min(ans, (1ll) * to * a[p[k]][i]);
}
}
}
return dp[n][k] = ans;
}
int main(){
scanf("%d",&n);
for(int i = 0; i < 9; i++){
a[p[i]][0] = 1;
while((long long) a[p[i]][pw[p[i]]] < INF / p[i]){
pw[p[i]]++;
a[p[i]][pw[p[i]]] = (1ll) * a[p[i]][pw[p[i]] - 1] * p[i];
}
}
long long ans = f(n,8);
printf("%lld \n",ans);
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