#include <bits/stdc++.h>
using namespace std;
int n;
long long lst, ans, cur;
map<long long, long long> m;
void maxi(long long &cur, long long b) {
  if (cur < b) cur = b;
}
int main() {
  scanf("%d%lld", &n, &lst);
  m[lst - 1] = 0;
  for (int i = 1; i < n; ++i) {
    scanf("%lld", &cur);
    if (cur < lst)
      for (auto p = m.lower_bound(lst = cur); p != m.end(); m.erase(p++)) {
        maxi(m[cur - 1], p->second + (p->first - cur + 1) / cur * cur * i);
        maxi(m[p->first % cur], p->second + p->first / cur * cur * i);
      }
  }
  for (auto &x : m) maxi(ans, x.second + x.first * n);
  printf("%lld", ans);
}

### Prompt

Please create a solution in cpp to the following problem:

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 n;
long long lst, ans, cur;
map<long long, long long> m;
void maxi(long long &cur, long long b) {
  if (cur < b) cur = b;
}
int main() {
  scanf("%d%lld", &n, &lst);
  m[lst - 1] = 0;
  for (int i = 1; i < n; ++i) {
    scanf("%lld", &cur);
    if (cur < lst)
      for (auto p = m.lower_bound(lst = cur); p != m.end(); m.erase(p++)) {
        maxi(m[cur - 1], p->second + (p->first - cur + 1) / cur * cur * i);
        maxi(m[p->first % cur], p->second + p->first / cur * cur * i);
      }
  }
  for (auto &x : m) maxi(ans, x.second + x.first * n);
  printf("%lld", ans);
}

```