#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
int n;
long long ans, a[maxn];
map<long long, long long> mp;
void chk(long long &x, long long y) {
  if (x < y) x = y;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  mp[a[1] - 1] = 0;
  long long lst = a[1];
  for (int i = 2; i <= n; i++)
    if (a[i] < lst) {
      while (1) {
        auto it = mp.lower_bound(a[i]);
        if (it == mp.end()) break;
        long long x = it->first, s = it->second;
        mp.erase(it);
        chk(mp[x % a[i]], s + (i - 1) * (x - x % a[i]));
        chk(mp[a[i] - 1], s + (i - 1) * (x - x % a[i] - a[i]));
      }
      lst = a[i];
    }
  for (auto p : mp) {
    chk(ans, p.first * n + p.second);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please formulate 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;
const int maxn = 200010;
int n;
long long ans, a[maxn];
map<long long, long long> mp;
void chk(long long &x, long long y) {
  if (x < y) x = y;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  mp[a[1] - 1] = 0;
  long long lst = a[1];
  for (int i = 2; i <= n; i++)
    if (a[i] < lst) {
      while (1) {
        auto it = mp.lower_bound(a[i]);
        if (it == mp.end()) break;
        long long x = it->first, s = it->second;
        mp.erase(it);
        chk(mp[x % a[i]], s + (i - 1) * (x - x % a[i]));
        chk(mp[a[i] - 1], s + (i - 1) * (x - x % a[i] - a[i]));
      }
      lst = a[i];
    }
  for (auto p : mp) {
    chk(ans, p.first * n + p.second);
  }
  printf("%lld\n", ans);
  return 0;
}

```