#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
long long n, ans, v;
map<long long, long long> f;
map<long long, long long>::iterator it;
inline long long rd() {
  long long ret = 0, f = 1;
  char gc = getchar();
  while (gc < '0' || gc > '9') {
    if (gc == '-') f = -f;
    gc = getchar();
  }
  while (gc >= '0' && gc <= '9') ret = ret * 10 + (gc ^ '0'), gc = getchar();
  return ret * f;
}
int main() {
  n = rd();
  long long i, a, b;
  f[rd() - 1] = 0;
  for (i = 2; i <= n; i++) {
    v = rd();
    while (f.begin() != f.end()) {
      it = f.end(), it--, 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));
      f.erase(it);
    }
  }
  for (it = f.begin(); it != f.end(); it++)
    ans = max(ans, n * ((*it).first) + (*it).second);
  printf("%I64d", 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;
const int maxn = 200010;
long long n, ans, v;
map<long long, long long> f;
map<long long, long long>::iterator it;
inline long long rd() {
  long long ret = 0, f = 1;
  char gc = getchar();
  while (gc < '0' || gc > '9') {
    if (gc == '-') f = -f;
    gc = getchar();
  }
  while (gc >= '0' && gc <= '9') ret = ret * 10 + (gc ^ '0'), gc = getchar();
  return ret * f;
}
int main() {
  n = rd();
  long long i, a, b;
  f[rd() - 1] = 0;
  for (i = 2; i <= n; i++) {
    v = rd();
    while (f.begin() != f.end()) {
      it = f.end(), it--, 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));
      f.erase(it);
    }
  }
  for (it = f.begin(); it != f.end(); it++)
    ans = max(ans, n * ((*it).first) + (*it).second);
  printf("%I64d", ans);
  return 0;
}

```