#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 5;
map<long long, long long> q;
vector<pair<long long, long long> > f;
int main() {
  int n, i;
  long long x, p;
  scanf("%d", &n);
  scanf("%lld", &p);
  q[p - 1] = 0;
  for (i = 1; i < n; i++) {
    scanf("%lld", &x);
    if (x >= p) continue;
    f.clear();
    auto e = q.lower_bound(x);
    for (auto u = e; u != q.end(); u++) f.push_back(*u);
    q.erase(e, q.end());
    for (auto u : f) {
      q[x - 1] = max(q[x - 1], (u.first - x + 1) / x * x * i + u.second);
      q[u.first % x] = max(q[u.first % x], u.first / x * x * i + u.second);
    }
  }
  x = 0;
  for (auto e : q) x = max(x, e.first * n + e.second);
  printf("%lld\n", x);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 mod = 1e9 + 7;
const int maxn = 1e5 + 5;
map<long long, long long> q;
vector<pair<long long, long long> > f;
int main() {
  int n, i;
  long long x, p;
  scanf("%d", &n);
  scanf("%lld", &p);
  q[p - 1] = 0;
  for (i = 1; i < n; i++) {
    scanf("%lld", &x);
    if (x >= p) continue;
    f.clear();
    auto e = q.lower_bound(x);
    for (auto u = e; u != q.end(); u++) f.push_back(*u);
    q.erase(e, q.end());
    for (auto u : f) {
      q[x - 1] = max(q[x - 1], (u.first - x + 1) / x * x * i + u.second);
      q[u.first % x] = max(q[u.first % x], u.first / x * x * i + u.second);
    }
  }
  x = 0;
  for (auto e : q) x = max(x, e.first * n + e.second);
  printf("%lld\n", x);
  return 0;
}

```