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

这题让我们求做少玩多少局，利用二分的思想，每次找一个数，看能不能行，可以的话就rightt=mid-1; 不行的话就leftt=mid+1;

这样到最后就能得到临界的最小值了。

#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iomanip>
using namespace std;

typedef long long LL;
const int maxn=100005;
LL n,m;
LL a[maxn];

int check(LL temp)
{
    LL cnt=0;
    for(int i=0;i<n;i++)
    {
        if(temp<a[i])
            return 0;
        cnt+=(temp-a[i]);
    }
    if(temp<=cnt)
        return 1;
    else
        return 0;
}

int main()
{
    scanf("%I64d",&n);
    for(int i=0;i<n;i++)
        scanf("%I64d",&a[i]);
    LL leftt=0,rightt=1e10,mid;
    while(leftt<=rightt)
    {
        mid=(leftt+rightt)/2;
        if(check(mid))
            rightt=mid-1;
        else
            leftt=mid+1;
    }
    cout<<leftt<<endl;
    return 0;
}