#include <bits/stdc++.h>
using namespace std;
map<long long, long long> mp;
long long ans, A, B, nex;
long long n, a[1000007];
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", a + i);
  mp[(long long)a[1] - 1] = 0;
  for (int i = 1; i <= n; i++) {
    while (1) {
      auto it = mp.lower_bound(a[i]);
      if (it == mp.end() || it->first < a[i]) break;
      A = it->first, B = it->second;
      mp.erase(it);
      nex = A % a[i];
      mp[nex] = max(mp[nex], B + (i - 1) * (A - A % a[i]));
      mp[a[i] - 1] = max(mp[a[i] - 1], B + (i - 1) * (A - A % a[i] - a[i]));
    }
  }
  for (auto i : mp) ans = max(ans, i.first * n + i.second);
  printf("%lld\n", ans);
}

### Prompt

Please create a solution in CPP 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;
map<long long, long long> mp;
long long ans, A, B, nex;
long long n, a[1000007];
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", a + i);
  mp[(long long)a[1] - 1] = 0;
  for (int i = 1; i <= n; i++) {
    while (1) {
      auto it = mp.lower_bound(a[i]);
      if (it == mp.end() || it->first < a[i]) break;
      A = it->first, B = it->second;
      mp.erase(it);
      nex = A % a[i];
      mp[nex] = max(mp[nex], B + (i - 1) * (A - A % a[i]));
      mp[a[i] - 1] = max(mp[a[i] - 1], B + (i - 1) * (A - A % a[i] - a[i]));
    }
  }
  for (auto i : mp) ans = max(ans, i.first * n + i.second);
  printf("%lld\n", ans);
}

```