/*
    Programmer : Alexandru Olteanu
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// GCC Optimizations
// #pragma GCC optimize("Ofast")
// #pragma GCC target("fma,sse,sse2,sse3,ssse3,sse4,popcnt")
// #pragma GCC target("abm,mmx,avx,avx2,tune=native")
// #pragma GCC optimize(3)
// #pragma GCC optimize("inline")
// #pragma GCC optimize("-fgcse")
// #pragma GCC optimize("-fgcse-lm")
// #pragma GCC optimize("-fipa-sra")
// #pragma GCC optimize("-ftree-pre")
// #pragma GCC optimize("-ftree-vrp")
// #pragma GCC optimize("-fpeephole2")
// #pragma GCC optimize("-ffast-math")
// #pragma GCC optimize("-fsched-spec")
// #pragma GCC optimize("unroll-loops")
// Useful
mt19937 rng((unsigned int) chrono::steady_clock::now().time_since_epoch().count());
#define FastEverything  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define HighPrecision cout << fixed << setprecision(17);
typedef long long ll;
typedef pair<int, int> pii;
const int mod = 1000000007;
const int mod2 = 100000000;
const int md = 998244353LL;
ll mypowr(ll a, ll b, ll mod1) {ll res = 1; if(b < 0)b = 0; a %= mod1; assert(b >= 0);
for(; b; b >>= 1){if (b & 1) res = res * a % mod1;a = a * a % mod1;} return res;}
ll mypow(ll a, ll b) {ll res = 1; if(b < 0)b = 0;assert(b >= 0);
for(; b; b >>= 1){if(b & 1) res = res * a;a = a * a;} return res;}
#define pb push_back
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

ifstream fin("fact.in");
ofstream fout("fact.out");
#define cin fin
#define cout fout

const ll infll = 0x7fffffffffffffff;
const int inf = 0x7fffffff;
const int maxn = 2e5 + 5;



int main() {

    FastEverything
    HighPrecision
    int test = 1;
    // cin>>test;
    for (int tt = 1; tt <= test; ++tt) {

        int n;
        cin >> n;
        int lo = 1, hi = 1e9;
        int ans = inf;
        while (lo <= hi) {
            int mid = lo + (hi - lo) / 2;
            ll have = 0;
            int x = mid;
            while (x >= 5) {
                have += x / 5;
                x /= 5;
                if (have > n) break;
            }
            if (have >= n) {
                if (have == n)ans = mid;
                hi = mid - 1;
            }
            else {
                lo = mid + 1;
            }
        }

        if (ans == inf) ans = -1;

        cout << ans << '\n';
        
    }

    return 0;
}