#include <bits/stdc++.h>
using namespace std;
int n;
long long x[200010], f[200010];
int main() {
  int i, j;
  long long k, l;
  scanf("%d", &n);
  for (i = n; i > 0; i--) scanf("%lld", &x[i]);
  for (i = n - 1; i > 0; i--) x[i] = min(x[i], x[i + 1]);
  x[0] = 1;
  for (i = 1; i <= n; i++)
    for (k = x[i], l = 0; k; k %= x[j]) {
      j = lower_bound(x + 1, x + n, k) - x - 1;
      f[i] = max(f[i], f[j] + (long long)(n - j) * (k - k % x[j] - x[j]) + l);
      l += (long long)(n - j) * (k - k % x[j]);
    }
  printf("%lld\n", f[n]);
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
int n;
long long x[200010], f[200010];
int main() {
  int i, j;
  long long k, l;
  scanf("%d", &n);
  for (i = n; i > 0; i--) scanf("%lld", &x[i]);
  for (i = n - 1; i > 0; i--) x[i] = min(x[i], x[i + 1]);
  x[0] = 1;
  for (i = 1; i <= n; i++)
    for (k = x[i], l = 0; k; k %= x[j]) {
      j = lower_bound(x + 1, x + n, k) - x - 1;
      f[i] = max(f[i], f[j] + (long long)(n - j) * (k - k % x[j] - x[j]) + l);
      l += (long long)(n - j) * (k - k % x[j]);
    }
  printf("%lld\n", f[n]);
  return 0;
}

```