#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 0;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f ? x : -x;
}
map<long long, long long> f;
map<long long, long long>::iterator it;
int main() {
  int n = read();
  f[read() - 1] = 0;
  for (int i = 2; i <= n; ++i) {
    long long v = read();
    for (it = --f.end();; f.erase(it--)) {
      long long a = (*it).first, b = (*it).second;
      if (a < v) break;
      f[v - 1] = max(f[v - 1], b + (i - 1) * (a - a % v - v));
      f[a % v] = max(f[a % v], b + (i - 1) * (a - a % v));
    }
  }
  long long ans = 0;
  for (it = f.begin(); it != f.end(); ++it)
    ans = max(ans, n * (*it).first + (*it).second);
  cout << ans << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 0;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f ? x : -x;
}
map<long long, long long> f;
map<long long, long long>::iterator it;
int main() {
  int n = read();
  f[read() - 1] = 0;
  for (int i = 2; i <= n; ++i) {
    long long v = read();
    for (it = --f.end();; f.erase(it--)) {
      long long a = (*it).first, b = (*it).second;
      if (a < v) break;
      f[v - 1] = max(f[v - 1], b + (i - 1) * (a - a % v - v));
      f[a % v] = max(f[a % v], b + (i - 1) * (a - a % v));
    }
  }
  long long ans = 0;
  for (it = f.begin(); it != f.end(); ++it)
    ans = max(ans, n * (*it).first + (*it).second);
  cout << ans << endl;
  return 0;
}

```