#include <bits/stdc++.h>
using namespace std;
set<pair<long long, long long> > S;
vector<pair<long long, long long> > v;
int n;
long long a[200000 + 5];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  S.insert(make_pair(a[1] - 1, 0));
  for (int i = 2; i <= n; i++) {
    v.clear();
    while (!S.empty()) {
      auto p = S.end();
      p--;
      if (p->first < a[i]) break;
      v.push_back(*p);
      S.erase(p);
    }
    if (!v.empty()) {
      long long mx = 0;
      for (auto t : v) {
        mx = max(mx, t.second + (i - 1) * (t.first - a[i] - t.first % a[i]));
        S.insert(make_pair(t.first % a[i],
                           t.second + (i - 1) * (t.first - t.first % a[i])));
      }
      S.insert(make_pair(a[i] - 1, mx));
    }
  }
  long long ans = 0;
  for (auto p : S) ans = max(ans, n * p.first + p.second);
  printf("%lld\n", ans);
}

### 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;
set<pair<long long, long long> > S;
vector<pair<long long, long long> > v;
int n;
long long a[200000 + 5];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  S.insert(make_pair(a[1] - 1, 0));
  for (int i = 2; i <= n; i++) {
    v.clear();
    while (!S.empty()) {
      auto p = S.end();
      p--;
      if (p->first < a[i]) break;
      v.push_back(*p);
      S.erase(p);
    }
    if (!v.empty()) {
      long long mx = 0;
      for (auto t : v) {
        mx = max(mx, t.second + (i - 1) * (t.first - a[i] - t.first % a[i]));
        S.insert(make_pair(t.first % a[i],
                           t.second + (i - 1) * (t.first - t.first % a[i])));
      }
      S.insert(make_pair(a[i] - 1, mx));
    }
  }
  long long ans = 0;
  for (auto p : S) ans = max(ans, n * p.first + p.second);
  printf("%lld\n", ans);
}

```