#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
map<long long, long long> dp;
int n;
long long a[N];
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) scanf("%lld", a + i);
  a[n + 1] = 1;
  dp[a[1] - 1] = 0;
  for (int i = (1); i <= (n); ++i)
    for (auto j = dp.lower_bound(a[i + 1]); j != dp.end(); dp.erase(j++)) {
      long long x = j->first;
      dp[x % a[i + 1]] =
          max(dp[x % a[i + 1]], j->second + (x - x % a[i + 1]) * i);
      dp[a[i + 1] - 1] =
          max(dp[a[i + 1] - 1],
              j->second + (x - a[i + 1] + 1) / a[i + 1] * a[i + 1] * i);
    }
  printf("%lld\n", dp[0]);
  return 0;
}

### Prompt

Generate 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 + 10;
map<long long, long long> dp;
int n;
long long a[N];
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) scanf("%lld", a + i);
  a[n + 1] = 1;
  dp[a[1] - 1] = 0;
  for (int i = (1); i <= (n); ++i)
    for (auto j = dp.lower_bound(a[i + 1]); j != dp.end(); dp.erase(j++)) {
      long long x = j->first;
      dp[x % a[i + 1]] =
          max(dp[x % a[i + 1]], j->second + (x - x % a[i + 1]) * i);
      dp[a[i + 1] - 1] =
          max(dp[a[i + 1] - 1],
              j->second + (x - a[i + 1] + 1) / a[i + 1] * a[i + 1] * i);
    }
  printf("%lld\n", dp[0]);
  return 0;
}

```