#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
int n;
long long Mx, ans, 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[a[1] - 1] = 0;
  Mx = a[1];
  for (int i = 1; i < n; i++) {
    for (map<long long, long long>::iterator j = mp.lower_bound(a[i + 1]);
         j != mp.end(); ++j) {
      long long x = j->first, y = j->second;
      if (x >= Mx) break;
      if (x >= a[i + 1]) {
        mp[x % a[i + 1]] =
            max(mp[x % a[i + 1]], y + 1ll * i * (x - x % a[i + 1]));
        mp[a[i + 1] - 1] =
            max(mp[a[i + 1] - 1],
                y + 1ll * i * ((x - a[i + 1] + 1) / a[i + 1] * a[i + 1]));
      }
    }
    Mx = min(Mx, a[i + 1]);
  }
  for (map<long long, long long>::iterator i = mp.begin(); i != mp.end(); ++i) {
    long long x = i->first, y = i->second;
    if (x >= Mx) break;
    ans = max(ans, y + x * n);
  }
  printf("%lld\n", ans);
  return 0;
}

### 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;
const int N = 200010;
int n;
long long Mx, ans, 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[a[1] - 1] = 0;
  Mx = a[1];
  for (int i = 1; i < n; i++) {
    for (map<long long, long long>::iterator j = mp.lower_bound(a[i + 1]);
         j != mp.end(); ++j) {
      long long x = j->first, y = j->second;
      if (x >= Mx) break;
      if (x >= a[i + 1]) {
        mp[x % a[i + 1]] =
            max(mp[x % a[i + 1]], y + 1ll * i * (x - x % a[i + 1]));
        mp[a[i + 1] - 1] =
            max(mp[a[i + 1] - 1],
                y + 1ll * i * ((x - a[i + 1] + 1) / a[i + 1] * a[i + 1]));
      }
    }
    Mx = min(Mx, a[i + 1]);
  }
  for (map<long long, long long>::iterator i = mp.begin(); i != mp.end(); ++i) {
    long long x = i->first, y = i->second;
    if (x >= Mx) break;
    ans = max(ans, y + x * n);
  }
  printf("%lld\n", ans);
  return 0;
}

```