#include <bits/stdc++.h>
using namespace std;
int n;
long long a;
map<long long, long long> M;
int main() {
  scanf("%d%lld", &n, &a);
  M[a - 1] = 0;
  for (int i = 2; i <= n; i++) {
    scanf("%lld", &a);
    for (map<long long, long long>::iterator it = M.lower_bound(a), nxt;
         it != M.end(); it = nxt) {
      long long x = it->first, y = it->second;
      M[x % a] = max(M[x % a], y + (x - x % a) * (i - 1));
      M[a - 1] = max(M[a - 1], y + (x - a + 1) / a * a * (i - 1));
      nxt = it;
      nxt++;
      M.erase(it);
    }
  }
  long long Ans = 0;
  for (auto p : M) Ans = max(Ans, p.first * n + p.second);
  printf("%lld\n", Ans);
}

### 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;
int n;
long long a;
map<long long, long long> M;
int main() {
  scanf("%d%lld", &n, &a);
  M[a - 1] = 0;
  for (int i = 2; i <= n; i++) {
    scanf("%lld", &a);
    for (map<long long, long long>::iterator it = M.lower_bound(a), nxt;
         it != M.end(); it = nxt) {
      long long x = it->first, y = it->second;
      M[x % a] = max(M[x % a], y + (x - x % a) * (i - 1));
      M[a - 1] = max(M[a - 1], y + (x - a + 1) / a * a * (i - 1));
      nxt = it;
      nxt++;
      M.erase(it);
    }
  }
  long long Ans = 0;
  for (auto p : M) Ans = max(Ans, p.first * n + p.second);
  printf("%lld\n", Ans);
}

```