#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int n;
long long A[N];
map<long long, long long> MP;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &A[i]);
  MP.insert({A[1] - 1, 0});
  for (int i = 2; i <= n; i++) {
    long long B = -1;
    while (MP.rbegin()->first >= A[i]) {
      long long a = MP.rbegin()->first;
      long long b = MP.rbegin()->second;
      MP[a % A[i]] = max(MP[a % A[i]], b + a / A[i] * A[i] * (i - 1));
      B = max(B, b + (a / A[i] - 1) * A[i] * (i - 1));
      MP.erase(a);
    }
    if (B >= 0) MP[A[i] - 1] = max(MP[A[i] - 1], B);
  }
  long long MX = 0;
  for (auto X : MP) MX = max(MX, X.first * n + X.second);
  return !printf("%lld\n", MX);
}

### Prompt

Develop a solution in cpp 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;
const int N = 200005;
int n;
long long A[N];
map<long long, long long> MP;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &A[i]);
  MP.insert({A[1] - 1, 0});
  for (int i = 2; i <= n; i++) {
    long long B = -1;
    while (MP.rbegin()->first >= A[i]) {
      long long a = MP.rbegin()->first;
      long long b = MP.rbegin()->second;
      MP[a % A[i]] = max(MP[a % A[i]], b + a / A[i] * A[i] * (i - 1));
      B = max(B, b + (a / A[i] - 1) * A[i] * (i - 1));
      MP.erase(a);
    }
    if (B >= 0) MP[A[i] - 1] = max(MP[A[i] - 1], B);
  }
  long long MX = 0;
  for (auto X : MP) MX = max(MX, X.first * n + X.second);
  return !printf("%lld\n", MX);
}

```