#include <bits/stdc++.h>
using namespace std;
int n;
long long u;
map<long long, long long> ma;
int main() {
  scanf("%d%lld", &n, &u);
  ma[u - 1] = 0;
  for (int i = 1; i < n; i++) {
    scanf("%lld", &u);
    while (ma.rbegin()->first >= u) {
      pair<long long, long long> cur = *ma.rbegin();
      ma.erase(cur.first);
      long long lst = cur.first / u * u;
      ma[cur.first - lst] = max(ma[cur.first - lst], lst * i + cur.second);
      ma[u - 1] = max(ma[u - 1], (lst - u) * i + cur.second);
    }
  }
  long long ans = 0;
  for (const pair<long long, long long> &u : ma)
    ans = max(ans, u.first * n + u.second);
  printf("%lld", ans);
}

### 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;
int n;
long long u;
map<long long, long long> ma;
int main() {
  scanf("%d%lld", &n, &u);
  ma[u - 1] = 0;
  for (int i = 1; i < n; i++) {
    scanf("%lld", &u);
    while (ma.rbegin()->first >= u) {
      pair<long long, long long> cur = *ma.rbegin();
      ma.erase(cur.first);
      long long lst = cur.first / u * u;
      ma[cur.first - lst] = max(ma[cur.first - lst], lst * i + cur.second);
      ma[u - 1] = max(ma[u - 1], (lst - u) * i + cur.second);
    }
  }
  long long ans = 0;
  for (const pair<long long, long long> &u : ma)
    ans = max(ans, u.first * n + u.second);
  printf("%lld", ans);
}

```