#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e+5 + 10;
long long n, a[maxn];
map<long long, long long> f;
map<long long, long long>::iterator fi, tfi;
int main() {
  long long i, ans = 0;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  f[a[0] - 1] = 0;
  for (i = 1; i < n; i++) {
    for (fi = f.end(), fi--; fi->first >= a[i];) {
      f[fi->first % a[i]] =
          max(f[fi->first % a[i]], fi->second + fi->first / a[i] * a[i] * i);
      f[a[i] - 1] =
          max(f[a[i] - 1], fi->second + (fi->first / a[i] - 1) * a[i] * i);
      tfi = fi--;
      f.erase(tfi);
    }
  }
  for (fi = f.begin(); fi != f.end(); fi++)
    ans = max(ans, fi->second + fi->first * n);
  cout << ans << endl;
  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;
const int maxn = 2e+5 + 10;
long long n, a[maxn];
map<long long, long long> f;
map<long long, long long>::iterator fi, tfi;
int main() {
  long long i, ans = 0;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  f[a[0] - 1] = 0;
  for (i = 1; i < n; i++) {
    for (fi = f.end(), fi--; fi->first >= a[i];) {
      f[fi->first % a[i]] =
          max(f[fi->first % a[i]], fi->second + fi->first / a[i] * a[i] * i);
      f[a[i] - 1] =
          max(f[a[i] - 1], fi->second + (fi->first / a[i] - 1) * a[i] * i);
      tfi = fi--;
      f.erase(tfi);
    }
  }
  for (fi = f.begin(); fi != f.end(); fi++)
    ans = max(ans, fi->second + fi->first * n);
  cout << ans << endl;
  return 0;
}

```