#include <bits/stdc++.h>
long long n, ans, a[200010];
std::map<long long, long long> f;
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", a + i);
  f[a[1] - 1] = 0;
  for (int i = 2; i <= n; i++)
    for (std::map<long long, long long>::iterator it = f.lower_bound(a[i]);
         it != f.end(); it = f.erase(it)) {
      long long x = it->first, y = it->second;
      f[x % a[i]] = std::max(f[x % a[i]], y + (i - 1) * (x - x % a[i]));
      f[a[i] - 1] =
          std::max(f[a[i] - 1], y + (i - 1) * ((x + 1) / a[i] * a[i] - a[i]));
    }
  for (std::map<long long, long long>::iterator it = f.begin(); it != f.end();
       ++it) {
    long long x = it->first, y = it->second;
    ans = std::max(ans, x * n + y);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
long long n, ans, a[200010];
std::map<long long, long long> f;
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", a + i);
  f[a[1] - 1] = 0;
  for (int i = 2; i <= n; i++)
    for (std::map<long long, long long>::iterator it = f.lower_bound(a[i]);
         it != f.end(); it = f.erase(it)) {
      long long x = it->first, y = it->second;
      f[x % a[i]] = std::max(f[x % a[i]], y + (i - 1) * (x - x % a[i]));
      f[a[i] - 1] =
          std::max(f[a[i] - 1], y + (i - 1) * ((x + 1) / a[i] * a[i] - a[i]));
    }
  for (std::map<long long, long long>::iterator it = f.begin(); it != f.end();
       ++it) {
    long long x = it->first, y = it->second;
    ans = std::max(ans, x * n + y);
  }
  printf("%lld\n", ans);
  return 0;
}

```