Mafia

Description

One day n friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other n - 1people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the i-th person wants to play a_i rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?

Input

The first line contains integer n(3 ≤ n ≤ 10⁵). The second line contains n space-separated integers a₁, a₂, ..., a_n(1 ≤ a_i ≤ 10⁹) — the i-th number in the list is the number of rounds the i-th person wants to play.

Output

In a single line print a single integer — the minimum number of game rounds the friends need to let the i-th person play at least a_irounds.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Sample Input

Input

33 2 2

Output

4

Input

42 2 2 2

Output

3

Hint

You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times:http://en.wikipedia.org/wiki/Mafia_(party_game).

 

解题思路：

先找到最大的数据，最小的局数肯定不会小于这个最大值，假设当前局数就是这个最大值，因为最大值减去每个人的数值就是每个人可以当裁判的次数，然后将每个人可以当裁判的次数相加如果大于等于最小局数，则当前局数是正确的，如果小于则另局数加一，而当裁判的总次数+n，然后将两个值再进行比较，

#include <iostream>#include<queue>using namespace std;int main(){    long long int n,i;    while(cin>>n)    {        long long int a[100099];        long long int max=0,sum=0;        for(i=0;i<n;i++)        {            cin>>a[i];            if(a[i]>max) max=a[i];        }        for(i=0;i<n;i++)        {            a[i]=max-a[i];            sum+=a[i];        }        for(i=0;i<1999999999;i++)        {            if(sum>=max){ cout<<max; break;}            else            {                sum+=n;                max++;            }        }    }    return 0;}