#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n;
long long mods[N];
long long tmp[N];
map<long long, long long, greater<long long>> dp;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> mods[i];
  }
  dp[mods[1] - 1] = 0;
  for (int i = 1; i < n; i++) {
    int cnt = 0;
    for (auto data : dp) {
      long long j = data.first;
      long long b = data.second;
      if (j < mods[i + 1]) {
        break;
      }
      dp[mods[i + 1] - 1] =
          max(dp[mods[i + 1] - 1],
              b + (long long)i *
                      (mods[i + 1] * ((j + 1) / mods[i + 1]) - mods[i + 1]));
      dp[j % mods[i + 1]] =
          max(dp[j % mods[i + 1]], (b + (long long)i * (j - j % mods[i + 1])));
      tmp[cnt++] = j;
    }
    while (cnt--) {
      dp.erase(tmp[cnt]);
    }
  }
  long long ans = 0;
  long long mmax = 0;
  for (auto i : dp) {
    ans = max(ans, i.first * n + i.second);
  }
  cout << ans << endl;
  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 + 5;
int n;
long long mods[N];
long long tmp[N];
map<long long, long long, greater<long long>> dp;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> mods[i];
  }
  dp[mods[1] - 1] = 0;
  for (int i = 1; i < n; i++) {
    int cnt = 0;
    for (auto data : dp) {
      long long j = data.first;
      long long b = data.second;
      if (j < mods[i + 1]) {
        break;
      }
      dp[mods[i + 1] - 1] =
          max(dp[mods[i + 1] - 1],
              b + (long long)i *
                      (mods[i + 1] * ((j + 1) / mods[i + 1]) - mods[i + 1]));
      dp[j % mods[i + 1]] =
          max(dp[j % mods[i + 1]], (b + (long long)i * (j - j % mods[i + 1])));
      tmp[cnt++] = j;
    }
    while (cnt--) {
      dp.erase(tmp[cnt]);
    }
  }
  long long ans = 0;
  long long mmax = 0;
  for (auto i : dp) {
    ans = max(ans, i.first * n + i.second);
  }
  cout << ans << endl;
  return 0;
}

```