#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i = (a), _b = (b); i <= _b; i++)
#define FORD(i, b, a) for (int i = (b), _a = (a); i >= _a; i--)
#define REP(i, n) for (int i = 0, _n = (n); i < _n; i++)
#define MAX 100005
int parent[MAX];
int n, Max = 0;
struct menkh {
int u, v, w;
};
vector<menkh> adj;
int root(int v) {
return v == parent[v] ? v : parent[v] = root(parent[v]);
}
bool unite(int u, int v) {
u = root(u), v = root(v);
if (u == v) return false;
parent[v] = u;
return true;
}
void solve() {
scanf("%d", &n);
vector<int> p(n + 1);
FOR(i, 1, n) {
scanf("%d", &p[i]);
Max = max(p[i], Max);
}
sort(p.begin() + 1, p.end());
p.erase(unique(p.begin() + 1, p.end()), p.end());
n = p.size() - 1;
FOR(i, 1, n) parent[i] = i;
FOR(i, 1, n) {
for (int k = 1; k * p[i] <= Max; k++) {
int v = lower_bound(p.begin() + 1, p.begin() + n + 1, k * p[i]) - p.begin();
if (v == i) v = i + 1;
if (v > n) break;
if (p[v] <= (k + 1) * p[i]) adj.push_back({i, v, p[v] % p[i]});
}
}
sort(adj.begin(), adj.end(), [](menkh A, menkh B) {
return A.w < B.w;
});
long long answer = 0;
REP(i, (int)adj.size()) {
if (unite(adj[i].u, adj[i].v)) answer += adj[i].w;
}
printf("%lld\n", answer);
}
int main() {
solve();
}
