#include <bits/stdc++.h>
using namespace std;
void m(long long &x, long long y) { x = max(x, y); }
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  long long ans = 0;
  cin >> n;
  map<long long, long long> dp;
  dp[1e13 + 100] = 0;
  for (int i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    auto point = dp.begin();
    while ((point = dp.lower_bound(x)) != dp.end()) {
      long long r = point->first, k = point->second;
      m(dp[r % x], k + (r - r % x) * i);
      m(dp[x - 1], k + (r - r % x - x) * i);
      dp.erase(point);
    }
  }
  for (auto point : dp) m(ans, point.first * n + point.second);
  cout << 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;
void m(long long &x, long long y) { x = max(x, y); }
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  long long ans = 0;
  cin >> n;
  map<long long, long long> dp;
  dp[1e13 + 100] = 0;
  for (int i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    auto point = dp.begin();
    while ((point = dp.lower_bound(x)) != dp.end()) {
      long long r = point->first, k = point->second;
      m(dp[r % x], k + (r - r % x) * i);
      m(dp[x - 1], k + (r - r % x - x) * i);
      dp.erase(point);
    }
  }
  for (auto point : dp) m(ans, point.first * n + point.second);
  cout << ans;
  return 0;
}

```