#include <bits/stdc++.h>
using namespace std;
set<pair<long long, long long> > s;
vector<pair<long long, long long> > g;
int n, m, l, k;
long long ans, x, lst;
int main() {
  scanf("%d%lld", &n, &lst);
  s.insert(make_pair(lst - 1, 0));
  for (int i = 1; i < n; i++) {
    scanf("%lld", &x);
    if (x >= lst) continue;
    lst = x;
    g.clear();
    while (!s.empty()) {
      auto it = s.end();
      it--;
      if ((*it).first < x) break;
      g.push_back((*it));
      s.erase(it);
    }
    if (!g.size()) continue;
    long long mx = 0;
    for (int j = 0; j < g.size(); j++) {
      pair<long long, long long> t = g[j];
      mx = max(mx, t.second + 1ll * i * (t.first - x - t.first % x));
      s.insert(
          make_pair(t.first % x, t.second + 1ll * i * (t.first - t.first % x)));
    }
    s.insert(make_pair(x - 1, mx));
  }
  for (auto it = s.begin(); it != s.end(); it++) {
    ans = max(ans, n * it->first + it->second);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

You are given a sequence of integers a1, a2, ..., an. Let <image>, and <image> for 1 ≤ i < n. Here, <image> denotes the modulus operation. Find the maximum value of f(x, 1) over all nonnegative integers x. 

Input

The first line contains a single integer n (1 ≤ n ≤ 200000) — the length of the sequence.

The second lines contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1013) — the elements of the sequence.

Output

Output a single integer — the maximum value of f(x, 1) over all nonnegative integers x.

Examples

Input

2
10 5


Output

13


Input

5
5 4 3 2 1


Output

6


Input

4
5 10 5 10


Output

16

Note

In the first example you can choose, for example, x = 19.

In the second example you can choose, for example, x = 3 or x = 2.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
set<pair<long long, long long> > s;
vector<pair<long long, long long> > g;
int n, m, l, k;
long long ans, x, lst;
int main() {
  scanf("%d%lld", &n, &lst);
  s.insert(make_pair(lst - 1, 0));
  for (int i = 1; i < n; i++) {
    scanf("%lld", &x);
    if (x >= lst) continue;
    lst = x;
    g.clear();
    while (!s.empty()) {
      auto it = s.end();
      it--;
      if ((*it).first < x) break;
      g.push_back((*it));
      s.erase(it);
    }
    if (!g.size()) continue;
    long long mx = 0;
    for (int j = 0; j < g.size(); j++) {
      pair<long long, long long> t = g[j];
      mx = max(mx, t.second + 1ll * i * (t.first - x - t.first % x));
      s.insert(
          make_pair(t.first % x, t.second + 1ll * i * (t.first - t.first % x)));
    }
    s.insert(make_pair(x - 1, mx));
  }
  for (auto it = s.begin(); it != s.end(); it++) {
    ans = max(ans, n * it->first + it->second);
  }
  printf("%lld\n", ans);
  return 0;
}

```