#include <bits/stdc++.h>
using namespace std;
map<long long, long long> f;
long long a[200010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  f[a[1] - 1] = 0;
  for (int i = 2; i <= n; i++)
    for (auto p = f.lower_bound(a[i]); p != f.end(); f.erase(p++)) {
      long long v = p->first;
      f[v % a[i]] = max(f[v % a[i]], p->second + (i - 1) * (v - v % a[i]));
      f[a[i] - 1] =
          max(f[a[i] - 1], p->second + (i - 1) * ((v + 1) / a[i] - 1) * a[i]);
    }
  long long ans = 0;
  for (auto p : f) ans = max(ans, p.first * n + p.second);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your task is to create 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;
map<long long, long long> f;
long long a[200010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  f[a[1] - 1] = 0;
  for (int i = 2; i <= n; i++)
    for (auto p = f.lower_bound(a[i]); p != f.end(); f.erase(p++)) {
      long long v = p->first;
      f[v % a[i]] = max(f[v % a[i]], p->second + (i - 1) * (v - v % a[i]));
      f[a[i] - 1] =
          max(f[a[i] - 1], p->second + (i - 1) * ((v + 1) / a[i] - 1) * a[i]);
    }
  long long ans = 0;
  for (auto p : f) ans = max(ans, p.first * n + p.second);
  printf("%lld\n", ans);
  return 0;
}

```