/*

 * Mafia.cpp

 *

 *  Created on: 2013-10-12

 *      Author: wangzhu

 */

/**

 * 每个人都想玩若干场，求至少需要玩几场才可以满足大家的需求。

 * 结果必然在某个人想玩的次数nmax（此人想玩的是最多的）与所有人想玩的次数和sum之间，

 * 故二分，left = nmax,right = sum,

 * 只需要需要玩的次数 * （总人数-1） >= 大家想玩的次数和即可

 *

 */

#include<cstdio>

#include<iostream>

using namespace std;

#define LL long long

#define NMAX 100010

int arr[NMAX];

LL binary(int n, int nmax, LL sum) {

    LL mid = -, left = nmax, right = sum;

    while (left <= right) {

        mid = left + (right - left) / ;

        if (sum <= (mid * n)) {

            right = mid - ;

        } else {

            left = mid + ;

        }

    }

    return left;

}

int main() {

    freopen("data.in", "r", stdin);

    int n, nmax;

    LL sum;

    while(~scanf("%d",&n)) {

        nmax = -;

        sum = ;

        for(int i = ;i < n;i++) {

            scanf("%d",arr + i);

            if(nmax < arr[i]) {

                nmax = arr[i];

            }

            sum += arr[i];

        }

        printf("%I64d\n",binary(n - ,nmax,sum));

    }

    return ;

}