#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  char ch = getchar();
  long long w = 1, c = 0;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') w = -1;
  for (; isdigit(ch); ch = getchar()) c = (c << 1) + (c << 3) + (ch ^ 48);
  return w * c;
}
const int M = 2e5 + 10;
int n;
long long a[M];
map<long long, long long> dp;
void upd(long long x, long long y) {
  dp.count(x) ? dp[x] = max(dp[x], y) : dp[x] = y;
}
int main() {
  n = read();
  for (int i = (1); i <= (n); ++i) {
    a[i] = read();
  }
  dp[a[1] - 1] = 0;
  for (int i = (2); i <= (n); ++i) {
    while (dp.size()) {
      auto O = *dp.rbegin();
      long long x = O.first, val = O.second;
      if (O.first >= a[i]) {
        long long nx = x % a[i], nval = val + (x - x % a[i]) * (i - 1);
        upd(nx, nval);
        nx = a[i] - 1;
        nval = val + (((x - a[i] + 1) / a[i]) * (a[i])) * (i - 1);
        upd(nx, nval);
        dp.erase(x);
      } else
        break;
    }
  }
  long long ans = 0;
  for (auto O : dp) {
    ans = max(ans, O.second + O.first * n);
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
inline long long read() {
  char ch = getchar();
  long long w = 1, c = 0;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') w = -1;
  for (; isdigit(ch); ch = getchar()) c = (c << 1) + (c << 3) + (ch ^ 48);
  return w * c;
}
const int M = 2e5 + 10;
int n;
long long a[M];
map<long long, long long> dp;
void upd(long long x, long long y) {
  dp.count(x) ? dp[x] = max(dp[x], y) : dp[x] = y;
}
int main() {
  n = read();
  for (int i = (1); i <= (n); ++i) {
    a[i] = read();
  }
  dp[a[1] - 1] = 0;
  for (int i = (2); i <= (n); ++i) {
    while (dp.size()) {
      auto O = *dp.rbegin();
      long long x = O.first, val = O.second;
      if (O.first >= a[i]) {
        long long nx = x % a[i], nval = val + (x - x % a[i]) * (i - 1);
        upd(nx, nval);
        nx = a[i] - 1;
        nval = val + (((x - a[i] + 1) / a[i]) * (a[i])) * (i - 1);
        upd(nx, nval);
        dp.erase(x);
      } else
        break;
    }
  }
  long long ans = 0;
  for (auto O : dp) {
    ans = max(ans, O.second + O.first * n);
  }
  cout << ans << "\n";
  return 0;
}

```