#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n;
  cin >> n;
  long long a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i > 0 && a[i] > a[i - 1]) a[i] = a[i - 1];
  }
  map<long long, long long> d;
  d[0] = 0;
  for (long long i = 0; i < n; i++) {
    long long best = 0;
    while (1) {
      long long v = -(*d.begin()).first;
      if (v < a[i]) break;
      long long p = d[-v];
      best = max(best, p + i * ((v + 1 - a[i]) - (v + 1) % a[i]));
      d.erase(-v);
      long long newv = v % a[i];
      d[-newv] = max(d[-newv], p + i * (v - newv));
    }
    d[-(a[i] - 1)] = max(d[-(a[i] - 1)], best);
  }
  long long ans = 0;
  for (auto it : d) {
    ans = max(ans, -n * (it.first) + it.second);
  }
  cout << ans << endl;
}

### 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;
int main() {
  long long n;
  cin >> n;
  long long a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i > 0 && a[i] > a[i - 1]) a[i] = a[i - 1];
  }
  map<long long, long long> d;
  d[0] = 0;
  for (long long i = 0; i < n; i++) {
    long long best = 0;
    while (1) {
      long long v = -(*d.begin()).first;
      if (v < a[i]) break;
      long long p = d[-v];
      best = max(best, p + i * ((v + 1 - a[i]) - (v + 1) % a[i]));
      d.erase(-v);
      long long newv = v % a[i];
      d[-newv] = max(d[-newv], p + i * (v - newv));
    }
    d[-(a[i] - 1)] = max(d[-(a[i] - 1)], best);
  }
  long long ans = 0;
  for (auto it : d) {
    ans = max(ans, -n * (it.first) + it.second);
  }
  cout << ans << endl;
}

```