#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
long long x[N], ans, a, b;
int n, m, k, j;
map<long long, long long> dp;
map<long long, long long>::iterator it;
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &x[i]);
  dp[x[1] - 1] = 0;
  x[n + 1] = 1;
  for (int i = 1; i < n; ++i) {
    while (1) {
      it = dp.lower_bound(x[i + 1]);
      if (it == dp.end()) break;
      a = it->first, b = it->second;
      dp.erase(a);
      dp[x[i + 1] - 1] =
          max(dp[x[i + 1] - 1], i * (a - a % x[i + 1] - x[i + 1]) + b);
      dp[a % x[i + 1]] = max(dp[a % x[i + 1]], i * (a - a % x[i + 1]) + b);
    }
  }
  for (it = dp.begin(); it != dp.end(); ++it) {
    ans = max(ans, dp[it->first] + (it->first) * n);
  }
  cout << ans;
}

### 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;
long long x[N], ans, a, b;
int n, m, k, j;
map<long long, long long> dp;
map<long long, long long>::iterator it;
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%lld", &x[i]);
  dp[x[1] - 1] = 0;
  x[n + 1] = 1;
  for (int i = 1; i < n; ++i) {
    while (1) {
      it = dp.lower_bound(x[i + 1]);
      if (it == dp.end()) break;
      a = it->first, b = it->second;
      dp.erase(a);
      dp[x[i + 1] - 1] =
          max(dp[x[i + 1] - 1], i * (a - a % x[i + 1] - x[i + 1]) + b);
      dp[a % x[i + 1]] = max(dp[a % x[i + 1]], i * (a - a % x[i + 1]) + b);
    }
  }
  for (it = dp.begin(); it != dp.end(); ++it) {
    ans = max(ans, dp[it->first] + (it->first) * n);
  }
  cout << ans;
}

```