#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
map<long long, long long>::iterator it;
map<long long, long long> f;
long long n, m, ans;
int main() {
  cin >> n >> m;
  f[m - 1] = 0;
  for (long long i = (2); i <= (n); i++) {
    cin >> m;
    it = f.lower_bound(m);
    while (it != f.end()) {
      f[it->first % m] = max(
          f[it->first % m], it->second + (i - 1) * (it->first - it->first % m));
      f[m - 1] =
          max(f[m - 1], it->second + (i - 1) * ((it->first + 1) / m * m - m));
      f.erase(it++);
    }
  }
  for (auto x : f) ans = max(ans, n * x.first + x.second);
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int maxn = 200005;
map<long long, long long>::iterator it;
map<long long, long long> f;
long long n, m, ans;
int main() {
  cin >> n >> m;
  f[m - 1] = 0;
  for (long long i = (2); i <= (n); i++) {
    cin >> m;
    it = f.lower_bound(m);
    while (it != f.end()) {
      f[it->first % m] = max(
          f[it->first % m], it->second + (i - 1) * (it->first - it->first % m));
      f[m - 1] =
          max(f[m - 1], it->second + (i - 1) * ((it->first + 1) / m * m - m));
      f.erase(it++);
    }
  }
  for (auto x : f) ans = max(ans, n * x.first + x.second);
  cout << ans << endl;
  return 0;
}

```