#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
long long a[maxn];
int n;
map<long long, long long> mp;
map<long long, long long>::iterator it;
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  mp[a[1] - 1] = 0;
  for (int i = 2; i <= n; ++i)
    for (it = mp.lower_bound(a[i]); it != mp.end();) {
      long long j = it->first;
      mp[j % a[i]] = max(mp[j % a[i]], mp[j] + (i - 1) * (j - j % a[i]));
      mp[a[i] - 1] = max(mp[a[i] - 1], mp[j] + (i - 1) * (j - j % a[i] - a[i]));
      mp.erase(it++);
    }
  long long ans = 0;
  for (it = mp.begin(); it != mp.end(); it++)
    ans = max(ans, n * it->first + it->second);
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int maxn = 2e5 + 10;
long long a[maxn];
int n;
map<long long, long long> mp;
map<long long, long long>::iterator it;
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  mp[a[1] - 1] = 0;
  for (int i = 2; i <= n; ++i)
    for (it = mp.lower_bound(a[i]); it != mp.end();) {
      long long j = it->first;
      mp[j % a[i]] = max(mp[j % a[i]], mp[j] + (i - 1) * (j - j % a[i]));
      mp[a[i] - 1] = max(mp[a[i] - 1], mp[j] + (i - 1) * (j - j % a[i] - a[i]));
      mp.erase(it++);
    }
  long long ans = 0;
  for (it = mp.begin(); it != mp.end(); it++)
    ans = max(ans, n * it->first + it->second);
  cout << ans << endl;
  return 0;
}

```