#include <bits/stdc++.h>
using namespace std;
int n;
long long ans;
map<long long, long long> f;
map<long long, long long>::iterator it;
template <class T>
inline void in(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9';) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  for (; ch >= '0' && ch <= '9';)
    x = x * 10 + (long long)ch - '0', ch = getchar();
  x = f ? -x : x;
}
int main() {
  long long v, a, b;
  in(n), in(v), f[v - 1] = 0;
  for (int i = 2; i <= n; ++i) {
    for (in(v); 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 + ((long long)i - 1) * (a - a % v - v));
      f[a % v] = max(f[a % v], b + ((long long)i - 1) * (a - a % v));
      f.erase(it);
    }
  }
  for (it = f.begin(); it != f.end(); ++it)
    ans = max(ans, 1LL * n * (*it).first + (*it).second);
  printf("%I64d\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>
using namespace std;
int n;
long long ans;
map<long long, long long> f;
map<long long, long long>::iterator it;
template <class T>
inline void in(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9';) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  for (; ch >= '0' && ch <= '9';)
    x = x * 10 + (long long)ch - '0', ch = getchar();
  x = f ? -x : x;
}
int main() {
  long long v, a, b;
  in(n), in(v), f[v - 1] = 0;
  for (int i = 2; i <= n; ++i) {
    for (in(v); 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 + ((long long)i - 1) * (a - a % v - v));
      f[a % v] = max(f[a % v], b + ((long long)i - 1) * (a - a % v));
      f.erase(it);
    }
  }
  for (it = f.begin(); it != f.end(); ++it)
    ans = max(ans, 1LL * n * (*it).first + (*it).second);
  printf("%I64d\n", ans);
  return 0;
}

```