#include <bits/stdc++.h>
using namespace std;
int n, i, j;
long long x, S, f[222222], a[222222];
int fd(long long x) {
  int l = 1, r = n + 1, md, an;
  if (x <= a[n]) return n + 1;
  for (; l <= r;)
    if (x > a[md = l + r >> 1])
      r = md - 1, an = md;
    else
      l = md + 1;
  return an;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  for (i = 2; i <= n; i++) a[i] = min(a[i], a[i - 1]);
  a[n + 1] = 1;
  for (i = n; i; i--)
    for (x = a[i], S = 0; x; x %= a[j]) {
      j = fd(x);
      f[i] = max(f[i], f[j] + (x - x % a[j] - a[j]) * (j - 1) + S);
      S += (x - x % a[j]) * (j - 1);
    }
  printf("%I64d", f[1]);
}

### 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;
int n, i, j;
long long x, S, f[222222], a[222222];
int fd(long long x) {
  int l = 1, r = n + 1, md, an;
  if (x <= a[n]) return n + 1;
  for (; l <= r;)
    if (x > a[md = l + r >> 1])
      r = md - 1, an = md;
    else
      l = md + 1;
  return an;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  for (i = 2; i <= n; i++) a[i] = min(a[i], a[i - 1]);
  a[n + 1] = 1;
  for (i = n; i; i--)
    for (x = a[i], S = 0; x; x %= a[j]) {
      j = fd(x);
      f[i] = max(f[i], f[j] + (x - x % a[j] - a[j]) * (j - 1) + S);
      S += (x - x % a[j]) * (j - 1);
    }
  printf("%I64d", f[1]);
}

```