#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
struct node {
  long long x, k;
  bool operator<(node r) const { return x == r.x ? k > r.k : x > r.x; }
};
set<node> q;
set<node>::iterator it;
long long a[N + 1];
int n;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &a[i]);
  q.insert((node){a[1] - 1, 0});
  for (int i = 1; i < n; ++i) {
    long long y = a[i + 1], mx = -1ll;
    while (1) {
      it = q.begin();
      if ((*it).x < y) break;
      long long x = (*it).x, k = (*it).k;
      mx = max(mx, k + (x - x % y - y) * i);
      q.insert((node){x % y, k + (x - x % y) * i}), q.erase(it);
    }
    if (~mx) q.insert((node){y - 1, mx});
  }
  long long ans = 0;
  for (it = q.begin(); it != q.end(); ++it)
    ans = max(ans, (*it).x * n + (*it).k);
  printf("%I64d\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 N = 2e5 + 5;
struct node {
  long long x, k;
  bool operator<(node r) const { return x == r.x ? k > r.k : x > r.x; }
};
set<node> q;
set<node>::iterator it;
long long a[N + 1];
int n;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%I64d", &a[i]);
  q.insert((node){a[1] - 1, 0});
  for (int i = 1; i < n; ++i) {
    long long y = a[i + 1], mx = -1ll;
    while (1) {
      it = q.begin();
      if ((*it).x < y) break;
      long long x = (*it).x, k = (*it).k;
      mx = max(mx, k + (x - x % y - y) * i);
      q.insert((node){x % y, k + (x - x % y) * i}), q.erase(it);
    }
    if (~mx) q.insert((node){y - 1, mx});
  }
  long long ans = 0;
  for (it = q.begin(); it != q.end(); ++it)
    ans = max(ans, (*it).x * n + (*it).k);
  printf("%I64d\n", ans);
  return 0;
}

```