#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
int n;
long long f[MAXN], a[MAXN];
int main() {
  scanf("%d", &n);
  for (int i = n; i; i--) scanf("%lld", &a[i]);
  for (int i = n - 1; i; i--) a[i] = min(a[i], a[i + 1]);
  a[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (long long j = a[i], k, l = 0; j; j %= a[k]) {
      k = lower_bound(a + 1, a + n + 1, j) - a - 1;
      f[i] = max(f[i], f[k] + (n - k) * (j - j % a[k] - a[k]) + l);
      l += (n - k) * (j - j % a[k]);
    }
  }
  printf("%lld\n", f[n]);
}

### 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 MAXN = 2e5 + 10;
int n;
long long f[MAXN], a[MAXN];
int main() {
  scanf("%d", &n);
  for (int i = n; i; i--) scanf("%lld", &a[i]);
  for (int i = n - 1; i; i--) a[i] = min(a[i], a[i + 1]);
  a[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (long long j = a[i], k, l = 0; j; j %= a[k]) {
      k = lower_bound(a + 1, a + n + 1, j) - a - 1;
      f[i] = max(f[i], f[k] + (n - k) * (j - j % a[k] - a[k]) + l);
      l += (n - k) * (j - j % a[k]);
    }
  }
  printf("%lld\n", f[n]);
}

```