E - Mafia CodeForces - 348A （推公式，思维）

E - Mafia CodeForces - 348A 

One day n friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other n - 1 people 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 ai 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, ..., an (1 ≤ ai ≤ 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 ai rounds.

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 %I64d specifier.

Examples

Input

3
3 2 2

Output

4

Input

4
2 2 2 2

Output

3

Note

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).

 

 题意：n个人玩游戏，第i人至少玩ai局，每局必须选出一个裁判，剩下的人才可以玩游戏，问一共至少要玩多少局游戏。

假设答案是ans，那么因为每局都要有一个裁判，所以ans*n-ans=sum（a）；所以ans=sum（a)/(n-1)向上取整，由于可能出现一个人要求局数特别大而其他人的局数特别小的情况，所以ans>=max(a),所以答案就是max（sum（a)/(n-1)向上取整，max(a)）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <queue>
#include <set>
#include <map>
#define loopa(i,a,b,d) for(int i=a;i<b;i+=d)
#define loops(i,a,b,d) for(int i=a;i>b;i-=d)
typedef long long LL;
using namespace std;
const int maxn=100000+1;
LL a[maxn];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        LL sum=0,maxx=0;
        loopa(i,0,n,1)
        {
            LL x;
            scanf("%I64d",&x);
            //printf("%I64d\n",x);
            maxx=max(maxx,x);
            a[i]=x;
            sum+=a[i];
        }
        LL ans=sum/(n-1);
        if(sum%(n-1)!=0)
        {
            ans++;
        }
        printf("%I64d\n",max(maxx,ans));
    }
    return 0;
}