#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n;
long long ans, a[N];
map<long long, long long> f;
map<long long, long long>::iterator it;
void chkmax(long long &x, long long y) {
  if (x < y) x = y;
}
int main() {
  cin >> n, a[n + 1] = 1;
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  f[a[1] - 1] = 0;
  for (int i = 1; i < n; ++i) {
    for (long long x, y;;) {
      it = f.lower_bound(a[i + 1]);
      if (it == f.end())
        break;
      else
        f.erase(it);
      x = it->first, y = it->second;
      chkmax(f[a[i + 1] - 1], 1ll * i * (x - x % a[i + 1] - a[i + 1]) + y);
      chkmax(f[x % a[i + 1]], 1ll * i * (x - x % a[i + 1]) + y);
    }
  }
  for (auto u : f) chkmax(ans, f[u.first] + u.first * n);
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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 N = 2e5 + 5;
int n;
long long ans, a[N];
map<long long, long long> f;
map<long long, long long>::iterator it;
void chkmax(long long &x, long long y) {
  if (x < y) x = y;
}
int main() {
  cin >> n, a[n + 1] = 1;
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  f[a[1] - 1] = 0;
  for (int i = 1; i < n; ++i) {
    for (long long x, y;;) {
      it = f.lower_bound(a[i + 1]);
      if (it == f.end())
        break;
      else
        f.erase(it);
      x = it->first, y = it->second;
      chkmax(f[a[i + 1] - 1], 1ll * i * (x - x % a[i + 1] - a[i + 1]) + y);
      chkmax(f[x % a[i + 1]], 1ll * i * (x - x % a[i + 1]) + y);
    }
  }
  for (auto u : f) chkmax(ans, f[u.first] + u.first * n);
  cout << ans << endl;
  return 0;
}

```