#include <bits/stdc++.h>
using namespace std;
const long long M = 200005;
long long read() {
  long long x = 0, f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9') {
    if (c == '-') f = -1;
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
long long n, ans, a[M];
map<long long, long long> mp;
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  mp[a[1] - 1] = 0;
  for (long long i = 2; i <= n; i++)
    for (map<long long, long long>::iterator it = mp.lower_bound(a[i]);
         it != mp.end(); mp.erase(it++)) {
      long long x = it->first, y = it->second;
      mp[x % a[i]] = max(mp[x % a[i]], y + (i - 1) * (x - x % a[i]));
      mp[a[i] - 1] =
          max(mp[a[i] - 1], y + (i - 1) * ((x + 1) / a[i] * a[i] - a[i]));
    }
  for (map<long long, long long>::iterator it = mp.begin(); it != mp.end();
       it++)
    ans = max(ans, it->second + n * it->first);
  printf("%lld\n", ans);
}

### Prompt

Create a solution in cpp for 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 long long M = 200005;
long long read() {
  long long x = 0, f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9') {
    if (c == '-') f = -1;
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
long long n, ans, a[M];
map<long long, long long> mp;
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  mp[a[1] - 1] = 0;
  for (long long i = 2; i <= n; i++)
    for (map<long long, long long>::iterator it = mp.lower_bound(a[i]);
         it != mp.end(); mp.erase(it++)) {
      long long x = it->first, y = it->second;
      mp[x % a[i]] = max(mp[x % a[i]], y + (i - 1) * (x - x % a[i]));
      mp[a[i] - 1] =
          max(mp[a[i] - 1], y + (i - 1) * ((x + 1) / a[i] * a[i] - a[i]));
    }
  for (map<long long, long long>::iterator it = mp.begin(); it != mp.end();
       it++)
    ans = max(ans, it->second + n * it->first);
  printf("%lld\n", ans);
}

```