#include <bits/stdc++.h>
using namespace std;
template <class T>
T smin(T &a, T b) {
  if (a > b) a = b;
  return a;
}
template <class T>
T smax(T &a, T b) {
  if (a < b) a = b;
  return a;
}
const int N = (int)(1e6) + 5;
long long s[N];
long long dp[N];
map<long long, long long> D;
int main(void) {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> s[i];
  for (int i = 1; i <= n; ++i) dp[i] = 0;
  for (int i = 2; i <= n; ++i) smin(s[i], s[i - 1]);
  s[++n] = 1;
  D[s[1] - 1] = 0;
  for (int i = 2; i <= n; ++i) {
    long long v = s[i];
    while (1) {
      auto it = D.lower_bound(s[i]);
      if (it == D.end()) break;
      long long x = it->first;
      long long y = it->second;
      D.erase(it);
      smax(D[x % v], (x - (x % v)) * (i - 1) + y);
      if (x >= v - 1) smax(D[v - 1], (x - (x % v) - v) * (i - 1) + y);
    }
  }
  cout << D[0] << '\n';
  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;
template <class T>
T smin(T &a, T b) {
  if (a > b) a = b;
  return a;
}
template <class T>
T smax(T &a, T b) {
  if (a < b) a = b;
  return a;
}
const int N = (int)(1e6) + 5;
long long s[N];
long long dp[N];
map<long long, long long> D;
int main(void) {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> s[i];
  for (int i = 1; i <= n; ++i) dp[i] = 0;
  for (int i = 2; i <= n; ++i) smin(s[i], s[i - 1]);
  s[++n] = 1;
  D[s[1] - 1] = 0;
  for (int i = 2; i <= n; ++i) {
    long long v = s[i];
    while (1) {
      auto it = D.lower_bound(s[i]);
      if (it == D.end()) break;
      long long x = it->first;
      long long y = it->second;
      D.erase(it);
      smax(D[x % v], (x - (x % v)) * (i - 1) + y);
      if (x >= v - 1) smax(D[v - 1], (x - (x % v) - v) * (i - 1) + y);
    }
  }
  cout << D[0] << '\n';
  return 0;
}

```