#include <bits/stdc++.h>
using namespace std;
const int N = 2e5;
mt19937_64 rng(std::chrono::system_clock::now().time_since_epoch().count());
int n;
long long a[N], ans, val;
vector<int> v;
map<long long, pair<long long, long long>> dp[N];
long long find(long long val) {
  int l = 0, r = v.size() - 1, bst = N;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (a[v[mid]] > val)
      l = mid + 1;
    else {
      bst = v[mid];
      r = mid - 1;
    }
  }
  return bst;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", a + i);
    if (v.empty() || a[v.back()] > a[i]) v.push_back(i);
  }
  dp[0][a[0] - 1] = {0, 0};
  for (int i = 0; i < n; i++)
    for (auto j : dp[i]) {
      int to = find(j.first);
      if (to == N) {
        val = max(val, j.second.first + n * j.first);
        if (val == j.second.first + n * j.first)
          ans = j.second.second + j.first;
        continue;
      }
      long long is = j.first, sum = j.second.first;
      long long x = is / a[to] * a[to], rem = is % a[to];
      dp[to][rem] = max(dp[to][rem], {sum + to * x, j.second.second + x});
      x -= a[to];
      rem = a[to] - 1;
      dp[to][rem] = max(dp[to][rem], {sum + to * x, j.second.second + x});
    }
  printf("%lld\n", val);
}

### 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 int N = 2e5;
mt19937_64 rng(std::chrono::system_clock::now().time_since_epoch().count());
int n;
long long a[N], ans, val;
vector<int> v;
map<long long, pair<long long, long long>> dp[N];
long long find(long long val) {
  int l = 0, r = v.size() - 1, bst = N;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (a[v[mid]] > val)
      l = mid + 1;
    else {
      bst = v[mid];
      r = mid - 1;
    }
  }
  return bst;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", a + i);
    if (v.empty() || a[v.back()] > a[i]) v.push_back(i);
  }
  dp[0][a[0] - 1] = {0, 0};
  for (int i = 0; i < n; i++)
    for (auto j : dp[i]) {
      int to = find(j.first);
      if (to == N) {
        val = max(val, j.second.first + n * j.first);
        if (val == j.second.first + n * j.first)
          ans = j.second.second + j.first;
        continue;
      }
      long long is = j.first, sum = j.second.first;
      long long x = is / a[to] * a[to], rem = is % a[to];
      dp[to][rem] = max(dp[to][rem], {sum + to * x, j.second.second + x});
      x -= a[to];
      rem = a[to] - 1;
      dp[to][rem] = max(dp[to][rem], {sum + to * x, j.second.second + x});
    }
  printf("%lld\n", val);
}

```