#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
map<long long, long long> mp;
int n;
long long num[N];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%lld", &num[i]);
  mp[num[1] - 1] = 0;
  for (int i = 1; i <= n; i++) {
    while (1) {
      auto it = mp.lower_bound(num[i]);
      if (it == mp.end() || (*it).first < num[i]) break;
      long long a = (*it).first, b = (*it).second;
      mp.erase(it);
      long long nex = a % num[i];
      mp[nex] = max(mp[nex], b + (i - 1) * (a - a % num[i]));
      mp[num[i] - 1] =
          max(mp[num[i] - 1], b + (i - 1) * (a - a % num[i] - num[i]));
    }
  }
  long long ans = 0;
  for (auto it : mp) {
    ans = max(ans, it.first * n + it.second);
  }
  cout << ans << endl;
  return 0;
}

### 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;
const int N = 2e5 + 100;
map<long long, long long> mp;
int n;
long long num[N];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%lld", &num[i]);
  mp[num[1] - 1] = 0;
  for (int i = 1; i <= n; i++) {
    while (1) {
      auto it = mp.lower_bound(num[i]);
      if (it == mp.end() || (*it).first < num[i]) break;
      long long a = (*it).first, b = (*it).second;
      mp.erase(it);
      long long nex = a % num[i];
      mp[nex] = max(mp[nex], b + (i - 1) * (a - a % num[i]));
      mp[num[i] - 1] =
          max(mp[num[i] - 1], b + (i - 1) * (a - a % num[i] - num[i]));
    }
  }
  long long ans = 0;
  for (auto it : mp) {
    ans = max(ans, it.first * n + it.second);
  }
  cout << ans << endl;
  return 0;
}

```