#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("%I64d", &a[i]);
  f[a[1] - 1] = 0;
  for (int i = 1; i < n; i++) {
    map<long long, long long>::iterator it = f.lower_bound(a[i + 1]);
    while (it != f.end()) {
      long long j = it->first;
      long long val = it->second;
      long long& to1 = f[a[i + 1] - 1];
      to1 = max(to1, val + (j - a[i + 1] + 1) / a[i + 1] * a[i + 1] * i);
      long long& to2 = f[j % a[i + 1]];
      to2 = max(to2, val + i * (j - j % a[i + 1]));
      f.erase(it++);
    }
  }
  long long ans = 0;
  for (map<long long, long long>::iterator it = f.begin(); it != f.end();
       it++) {
    long long j = it->first;
    long long val = it->second;
    ans = max(ans, val + n * j);
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Generate 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("%I64d", &a[i]);
  f[a[1] - 1] = 0;
  for (int i = 1; i < n; i++) {
    map<long long, long long>::iterator it = f.lower_bound(a[i + 1]);
    while (it != f.end()) {
      long long j = it->first;
      long long val = it->second;
      long long& to1 = f[a[i + 1] - 1];
      to1 = max(to1, val + (j - a[i + 1] + 1) / a[i + 1] * a[i + 1] * i);
      long long& to2 = f[j % a[i + 1]];
      to2 = max(to2, val + i * (j - j % a[i + 1]));
      f.erase(it++);
    }
  }
  long long ans = 0;
  for (map<long long, long long>::iterator it = f.begin(); it != f.end();
       it++) {
    long long j = it->first;
    long long val = it->second;
    ans = max(ans, val + n * j);
  }
  printf("%I64d\n", ans);
  return 0;
}

```