#include <bits/stdc++.h>
using namespace std;
long long n, ans, a[200005];
map<long long, long long> mp;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  mp.insert(pair<long long, long long>(a[1], 0));
  map<long long, long long>::iterator j;
  for (int i = 1; i < n; i++)
    for (j = mp.upper_bound(a[i + 1]); j != mp.end();) {
      long long &f0 = mp[a[i + 1]];
      f0 = max(f0, j->second + (j->first / a[i + 1] - 1) * a[i + 1] * i);
      if (j->first % a[i + 1]) {
        long long &f1 = mp[j->first % a[i + 1]];
        f1 = max(f1, j->second + j->first / a[i + 1] * a[i + 1] * i);
      }
      mp.erase(j);
      j = mp.upper_bound(a[i + 1]);
    }
  for (j = mp.begin(); j != mp.end(); j++)
    ans = max(ans, (j->first - 1) * n + j->second);
  printf("%I64d", ans);
}

### Prompt

Please formulate 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;
long long n, ans, a[200005];
map<long long, long long> mp;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  mp.insert(pair<long long, long long>(a[1], 0));
  map<long long, long long>::iterator j;
  for (int i = 1; i < n; i++)
    for (j = mp.upper_bound(a[i + 1]); j != mp.end();) {
      long long &f0 = mp[a[i + 1]];
      f0 = max(f0, j->second + (j->first / a[i + 1] - 1) * a[i + 1] * i);
      if (j->first % a[i + 1]) {
        long long &f1 = mp[j->first % a[i + 1]];
        f1 = max(f1, j->second + j->first / a[i + 1] * a[i + 1] * i);
      }
      mp.erase(j);
      j = mp.upper_bound(a[i + 1]);
    }
  for (j = mp.begin(); j != mp.end(); j++)
    ans = max(ans, (j->first - 1) * n + j->second);
  printf("%I64d", ans);
}

```