#include <bits/stdc++.h>
using namespace std;
inline long long Read() {
  long long x = 0;
  int f = 1, c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
long long ans, lst, cur;
map<long long, long long> a;
int n;
int main() {
  n = Read(), lst = Read(), a[lst - 1] = 0;
  for (int i = 1; i < n; i++)
    if ((cur = Read()) < lst)
      for (auto it = a.lower_bound(lst = cur); it != a.end(); a.erase(it++))
        a[cur - 1] =
            max(a[cur - 1], it->second + (it->first - cur + 1) / cur * cur * i),
                a[it->first % cur] = max(
                    a[it->first % cur], it->second + it->first / cur * cur * i);
  for (auto x : a) ans = max(ans, x.first * n + x.second);
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
inline long long Read() {
  long long x = 0;
  int f = 1, c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
long long ans, lst, cur;
map<long long, long long> a;
int n;
int main() {
  n = Read(), lst = Read(), a[lst - 1] = 0;
  for (int i = 1; i < n; i++)
    if ((cur = Read()) < lst)
      for (auto it = a.lower_bound(lst = cur); it != a.end(); a.erase(it++))
        a[cur - 1] =
            max(a[cur - 1], it->second + (it->first - cur + 1) / cur * cur * i),
                a[it->first % cur] = max(
                    a[it->first % cur], it->second + it->first / cur * cur * i);
  for (auto x : a) ans = max(ans, x.first * n + x.second);
  cout << ans << endl;
  return 0;
}

```