#include <bits/stdc++.h>
using namespace std;
const long long maxn = 200010;
long long a[maxn], ans, now, n;
set<pair<long long, long long> > S;
vector<pair<long long, long long> > V;
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  S.insert(make_pair(a[1] - 1, 0LL));
  now = a[1];
  for (long long i = 2; i <= n; i++) {
    if (a[i] >= now) continue;
    now = a[i];
    V.clear();
    while (!S.empty()) {
      set<pair<long long, long long> >::iterator it = S.end();
      it--;
      pair<long long, long long> temp = (*it);
      if (temp.first < a[i]) break;
      V.push_back(temp);
      S.erase(it);
    }
    if (V.empty()) continue;
    long long maxx = 0;
    for (long long j = 0; j < V.size(); j++) {
      pair<long long, long long> temp = V[j];
      maxx = max(maxx, temp.second +
                           (i - 1) * (temp.first - temp.first % a[i] - a[i]));
      S.insert(
          make_pair(temp.first % a[i],
                    temp.second + (i - 1) * (temp.first - temp.first % a[i])));
    }
    S.insert(make_pair(a[i] - 1, maxx));
  }
  for (set<pair<long long, long long> >::iterator it = S.begin(); it != S.end();
       it++) {
    pair<long long, long long> temp = (*it);
    ans = max(ans, temp.first * n + temp.second);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please formulate 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 long long maxn = 200010;
long long a[maxn], ans, now, n;
set<pair<long long, long long> > S;
vector<pair<long long, long long> > V;
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  S.insert(make_pair(a[1] - 1, 0LL));
  now = a[1];
  for (long long i = 2; i <= n; i++) {
    if (a[i] >= now) continue;
    now = a[i];
    V.clear();
    while (!S.empty()) {
      set<pair<long long, long long> >::iterator it = S.end();
      it--;
      pair<long long, long long> temp = (*it);
      if (temp.first < a[i]) break;
      V.push_back(temp);
      S.erase(it);
    }
    if (V.empty()) continue;
    long long maxx = 0;
    for (long long j = 0; j < V.size(); j++) {
      pair<long long, long long> temp = V[j];
      maxx = max(maxx, temp.second +
                           (i - 1) * (temp.first - temp.first % a[i] - a[i]));
      S.insert(
          make_pair(temp.first % a[i],
                    temp.second + (i - 1) * (temp.first - temp.first % a[i])));
    }
    S.insert(make_pair(a[i] - 1, maxx));
  }
  for (set<pair<long long, long long> >::iterator it = S.begin(); it != S.end();
       it++) {
    pair<long long, long long> temp = (*it);
    ans = max(ans, temp.first * n + temp.second);
  }
  printf("%lld\n", ans);
  return 0;
}

```