#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n;
long long a[N];
struct interval {
  long long l, r, b;
  bool operator<(const interval &x) const {
    if (r != x.r) return r > x.r;
    if (l != x.l) return l > x.l;
    return b > x.b;
  }
};
set<interval> S;
void insert(long long l, long long r, long long b) {
  S.insert((interval){l, r, b});
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 2; i <= n; i++) a[i] = min(a[i], a[i - 1]);
  insert(0, a[1] - 1, 0);
  for (int i = 2; i <= n; i++) {
    while (S.begin()->r >= a[i]) {
      interval now = *S.begin();
      S.erase(now);
      long long l = now.l, r = now.r, b = now.b;
      long long pl = l % a[i], pr = r % a[i];
      long long blo = r / a[i] * a[i];
      if (r - l + 1 >= a[i]) {
        insert(0, pr, b + blo * (i - 1));
        if (pr < a[i] - 1) insert(pr + 1, a[i] - 1, b + (blo - a[i]) * (i - 1));
      } else if (pl <= pr)
        insert(pl, pr, b + blo * (i - 1));
      else {
        insert(0, pr, b + blo * (i - 1));
        insert(pl, a[i] - 1, b + (blo - a[i]) * (i - 1));
      }
    }
  }
  long long ans = 0;
  for (auto I : S) ans = max(ans, I.r * n + I.b);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

In CPP, your task is to solve 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 N = 2e5 + 10;
int n;
long long a[N];
struct interval {
  long long l, r, b;
  bool operator<(const interval &x) const {
    if (r != x.r) return r > x.r;
    if (l != x.l) return l > x.l;
    return b > x.b;
  }
};
set<interval> S;
void insert(long long l, long long r, long long b) {
  S.insert((interval){l, r, b});
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 2; i <= n; i++) a[i] = min(a[i], a[i - 1]);
  insert(0, a[1] - 1, 0);
  for (int i = 2; i <= n; i++) {
    while (S.begin()->r >= a[i]) {
      interval now = *S.begin();
      S.erase(now);
      long long l = now.l, r = now.r, b = now.b;
      long long pl = l % a[i], pr = r % a[i];
      long long blo = r / a[i] * a[i];
      if (r - l + 1 >= a[i]) {
        insert(0, pr, b + blo * (i - 1));
        if (pr < a[i] - 1) insert(pr + 1, a[i] - 1, b + (blo - a[i]) * (i - 1));
      } else if (pl <= pr)
        insert(pl, pr, b + blo * (i - 1));
      else {
        insert(0, pr, b + blo * (i - 1));
        insert(pl, a[i] - 1, b + (blo - a[i]) * (i - 1));
      }
    }
  }
  long long ans = 0;
  for (auto I : S) ans = max(ans, I.r * n + I.b);
  printf("%lld\n", ans);
  return 0;
}

```