#include <bits/stdc++.h>
using namespace std;
struct Node {
  long long k, j;
  bool operator<(const Node& a) const { return j < a.j; }
};
priority_queue<Node> Q;
const int maxn = 2e5 + 7;
long long a[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  Q.push((Node){0, a[1] - 1});
  for (int i = 2; i <= n; i++) {
    long long Mx = 0;
    bool flag = 0;
    while (1) {
      Node u = Q.top();
      if (u.j < a[i]) break;
      flag = 1;
      Mx = max(Mx, u.k + a[i] * (i - 1) * ((u.j + 1) / a[i] - 1));
      Q.pop();
      Q.push((Node){u.k + 1LL * (i - 1) * (u.j - u.j % a[i]), u.j % a[i]});
    }
    if (flag) Q.push((Node){Mx, a[i] - 1});
  }
  long long ans = 0;
  while (Q.size()) {
    Node u = Q.top();
    Q.pop();
    ans = max(ans, u.k + u.j * n);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
struct Node {
  long long k, j;
  bool operator<(const Node& a) const { return j < a.j; }
};
priority_queue<Node> Q;
const int maxn = 2e5 + 7;
long long a[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  Q.push((Node){0, a[1] - 1});
  for (int i = 2; i <= n; i++) {
    long long Mx = 0;
    bool flag = 0;
    while (1) {
      Node u = Q.top();
      if (u.j < a[i]) break;
      flag = 1;
      Mx = max(Mx, u.k + a[i] * (i - 1) * ((u.j + 1) / a[i] - 1));
      Q.pop();
      Q.push((Node){u.k + 1LL * (i - 1) * (u.j - u.j % a[i]), u.j % a[i]});
    }
    if (flag) Q.push((Node){Mx, a[i] - 1});
  }
  long long ans = 0;
  while (Q.size()) {
    Node u = Q.top();
    Q.pop();
    ans = max(ans, u.k + u.j * n);
  }
  cout << ans << endl;
  return 0;
}

```