#include <bits/stdc++.h>
using namespace std;
long long rdn() {
  long long ret = 0;
  bool fx = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') fx = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') ret = ret * 10 + ch - '0', ch = getchar();
  return fx ? ret : -ret;
}
long long Mx(long long a, long long b) { return a > b ? a : b; }
const int N = 2e5 + 5;
int n;
long long a[N], ans;
map<long long, long long> mp;
map<long long, long long>::iterator it;
int main() {
  n = rdn();
  for (int i = 1; i <= n; i++) a[i] = rdn();
  mp[a[1] - 1] = 0;
  for (int i = 1; i < n; i++) {
    while (1) {
      it = mp.end();
      it--;
      long long r = (*it).first, k = (*it).second;
      if (r < a[i + 1]) break;
      mp.erase(it);
      mp[a[i + 1] - 1] =
          Mx(mp[a[i + 1] - 1], k + i * (r - r % a[i + 1] - a[i + 1]));
      mp[r % a[i + 1]] = Mx(mp[r % a[i + 1]], k + i * (r - r % a[i + 1]));
    }
  }
  for (it = mp.begin(); it != mp.end(); it++)
    ans = Mx(ans, n * (*it).first + (*it).second);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long rdn() {
  long long ret = 0;
  bool fx = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') fx = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') ret = ret * 10 + ch - '0', ch = getchar();
  return fx ? ret : -ret;
}
long long Mx(long long a, long long b) { return a > b ? a : b; }
const int N = 2e5 + 5;
int n;
long long a[N], ans;
map<long long, long long> mp;
map<long long, long long>::iterator it;
int main() {
  n = rdn();
  for (int i = 1; i <= n; i++) a[i] = rdn();
  mp[a[1] - 1] = 0;
  for (int i = 1; i < n; i++) {
    while (1) {
      it = mp.end();
      it--;
      long long r = (*it).first, k = (*it).second;
      if (r < a[i + 1]) break;
      mp.erase(it);
      mp[a[i + 1] - 1] =
          Mx(mp[a[i + 1] - 1], k + i * (r - r % a[i + 1] - a[i + 1]));
      mp[r % a[i + 1]] = Mx(mp[r % a[i + 1]], k + i * (r - r % a[i + 1]));
    }
  }
  for (it = mp.begin(); it != mp.end(); it++)
    ans = Mx(ans, n * (*it).first + (*it).second);
  printf("%lld\n", ans);
  return 0;
}

```