#include <bits/stdc++.h>
using namespace std;
map<long long, long long> f;
map<long long, long long>::iterator it;
int main() {
  long long n;
  scanf("%lld", &n);
  long long x;
  scanf("%lld", &x);
  f[x - 1] = 0;
  for (long long i = 1; i < n; i++) {
    long long x;
    scanf("%lld", &x);
    while (f.begin() != f.end()) {
      it = --f.end();
      long long a, b;
      a = (*it).first;
      b = (*it).second;
      if (a < x) break;
      f[x - 1] = max(f[x - 1], b + i * (a - a % x - x));
      f[a % x] = max(f[a % x], b + i * (a - a % x));
      f.erase(it);
    }
  }
  long long ans = 0;
  for (it = f.begin(); it != f.end(); it++)
    ans = max(ans, n * (*it).first + (*it).second);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
map<long long, long long>::iterator it;
int main() {
  long long n;
  scanf("%lld", &n);
  long long x;
  scanf("%lld", &x);
  f[x - 1] = 0;
  for (long long i = 1; i < n; i++) {
    long long x;
    scanf("%lld", &x);
    while (f.begin() != f.end()) {
      it = --f.end();
      long long a, b;
      a = (*it).first;
      b = (*it).second;
      if (a < x) break;
      f[x - 1] = max(f[x - 1], b + i * (a - a % x - x));
      f[a % x] = max(f[a % x], b + i * (a - a % x));
      f.erase(it);
    }
  }
  long long ans = 0;
  for (it = f.begin(); it != f.end(); it++)
    ans = max(ans, n * (*it).first + (*it).second);
  printf("%lld\n", ans);
  return 0;
}

```