#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
long long n, ans, t;
map<long long, long long> f;
map<long long, long long>::iterator res;
void read(long long &x) {
  x = 0;
  long long f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  x *= f;
}
int main() {
  long long i, a, b, c;
  read(n);
  read(c);
  f[c - 1] = 0;
  for (i = 2; i <= n; i++) {
    read(t);
    while (f.begin() != f.end()) {
      res = f.end(), res--, a = (*res).first, b = (*res).second;
      if (a < t) break;
      f[t - 1] = max(f[t - 1], b + (i - 1) * (a - a % t - t));
      f[a % t] = max(f[a % t], b + (i - 1) * (a - a % t));
      f.erase(res);
    }
  }
  for (res = f.begin(); res != f.end(); res++)
    ans = max(ans, n * ((*res).first) + (*res).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, t;
map<long long, long long> f;
map<long long, long long>::iterator res;
void read(long long &x) {
  x = 0;
  long long f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  x *= f;
}
int main() {
  long long i, a, b, c;
  read(n);
  read(c);
  f[c - 1] = 0;
  for (i = 2; i <= n; i++) {
    read(t);
    while (f.begin() != f.end()) {
      res = f.end(), res--, a = (*res).first, b = (*res).second;
      if (a < t) break;
      f[t - 1] = max(f[t - 1], b + (i - 1) * (a - a % t - t));
      f[a % t] = max(f[a % t], b + (i - 1) * (a - a % t));
      f.erase(res);
    }
  }
  for (res = f.begin(); res != f.end(); res++)
    ans = max(ans, n * ((*res).first) + (*res).second);
  printf("%I64d", ans);
  return 0;
}

```