Problem:
Score : 600 points

Problem Statement

We have a sequence of length N consisting of non-negative integers. Consider performing the following operation on this sequence until the largest element in this sequence becomes N-1 or smaller. (The operation is the same as the one in Problem D.)

Determine the largest element in the sequence (if there is more than one, choose one). Decrease the value of this element by N, and increase each of the other elements by 1.

It can be proved that the largest element in the sequence becomes N-1 or smaller after a finite number of operations.

You are given the sequence a_i. Find the number of times we will perform the above operation.

Constraints

2 ≤ N ≤ 50

0 ≤ a_i ≤ 10^{16} + 1000

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the number of times the operation will be performed.

Sample Input 1

4
3 3 3 3

Sample Output 1

0

Sample Input 2

3
1 0 3

Sample Output 2

1

Sample Input 3

2
2 2

Sample Output 3

2

Sample Input 4

7
27 0 0 0 0 0 0

Sample Output 4

3

Sample Input 5

10
1000 193 256 777 0 1 1192 1234567891011 48 425

Sample Output 5

1234567894848

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#ifdef MBP
#include "stdc++.h"
#else
#include <bits/stdc++.h>
#endif

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define REP(i, n) for (long long int i = 0, i##_len = (n); i < i##_len; ++i)
#define REP_R(i, n) for (long long int i = (n - 1); i >= 0; --i)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((long long int)(x).size())
#define BIT(n) (1LL << (n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());

#define INF INT_MAX
#define INFLL LLONG_MAX
#define PB push_back
#define MOD 1000000007LL
#define LSB(x) __builtin_ctzll(x)
#define MSB(x) 64LL - __builtin_clzll(x) - 1LL
#define MSBINT(x) 32LL - __builtin_clz(x)
#define BITCNT(x) __builtin_popcountll(x)

using namespace std;
using uif = uint_fast64_t;
using ll = long long int;
using vec = vector<ll>;
using mat = vector<vector<ll>>;
using pa = pair<ll, ll>;
using tTree = __gnu_pbds::tree<ll, __gnu_pbds::null_type, less<ll>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update>;

template <class T>
using PQ = priority_queue<T>;
template <class T>
using PQG = priority_queue<T, vector<T>, greater<T>>;

const ll dx[4] = {1, 0, -1, 0};
const ll dy[4] = {0, 1, 0, -1};
const ll dx8[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy8[8] = {0, 1, 1, 1, 0, -1, -1, -1};

template <class T>
bool chmax(T &, const T &);
template <class T>
bool chmin(T &, const T &);
ll gcd(ll, ll);
ll qp(ll a, ll b);
ll qp(ll a, ll b, ll mo);

ll mod_pow(ll, ll);
ll mod_add(ll, ll);
ll mod_mul(ll, ll);
ll mod_div(ll, ll);
void make_fact(ll);
void make_invfact(ll);
void init_fact();

ll combination(ll, ll);

#define FACSIZE 200002LL

ll invfac[FACSIZE];
ll fac[FACSIZE];

ll f(ll a, ll k, ll n)
{

  //return ki * (n + 1LL) >= (k + a + 1 - n);
  ll res = (-(n - 1LL) + (a + k) + n + 1LL - 1LL) / (n + 1LL);
  return max(0LL, res);
}

int main(void)
{
  ll n;
  cin >> n;
  vec A(n);
  REP(i, n)
  cin >> A[i];
  sort(begin(A), end(A), greater<ll>());

  /*
  for (auto x : A)
    cout << x << " ";
  cout << endl;
  //*/

  ll l = 0;
  ll r = LLONG_MAX / 51LL;

  while (r - l > 1)
  {
    //cout << l << "," << r << endl;
    ll piv = l + (r - l) / 2LL;

    bool bf = false;
    ll rest = piv;
    REP(i, n)
    {
      ll res = f(A[i], piv, n);
      //cout << res << endl;
      if (rest < res)
      {
        bf = true;
        break;
      }
      else
        rest -= res;
    }
    if (bf)
      l = piv;
    else
      r = piv;
  }
  ll ans = r;
  //cout << l << "," << r << endl;
  for (ll piv = r; piv >= r - 2 * n && piv >= 0; piv--)
  {
    bool bf = false;
    ll rest = piv;
    REP(i, n)
    {
      ll res = f(A[i], piv, n);
      //cout << res << endl;
      if (rest < res)
      {
        bf = true;
        break;
      }
      else
        rest -= res;
    }
    if (bf)
      continue;
    else
      ans = piv;
  }

  cout << ans << endl;

  return 0;
}

template <class T>
bool chmax(T &a, const T &b)
{
  if (a < b)
  {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b)
{
  if (b < a)
  {
    a = b;
    return 1;
  }
  return 0;
}

ll qp(ll a, ll b)
{
  ll ans = 1LL;
  do
  {
    if (b & 1LL)
      ans = 1LL * mod_mul(ans, a) % MOD;
    a = 1LL * mod_mul(a, a) % MOD;
  } while (b >>= 1LL);
  return ans;
}
ll qp(ll a, ll b, ll mo)
{
  ll ans = 1LL;
  do
  {
    if (b & 1LL)
      ans = 1LL * (ans % mo) * (a % mo);
    a = 1LL * (a % mo) * (a % mo);
  } while (b >>= 1LL);
  return ans;
}
ll gcd(ll a, ll b)
{
  return b ? gcd(b, a % b) : a;
}
ll mod_pow(ll x, ll r)
{
  if (r == 0)
    return 1;
  else if (r == 1)
    return x % MOD;
  else if (r % 2 == 0)
  {
    ll t = mod_pow(x, r / 2) % MOD;
    return t * t % MOD;
  }
  else
  {
    ll t = mod_pow(x, r / 2) % MOD;
    ll k = t * t % MOD;
    return k * x % MOD;
  }
}
ll mod_add(ll a, ll b)
{
  return (a + b) % MOD;
}

ll mod_mul(ll a, ll b)
{
  return (a * b) % MOD;
}

ll mod_div(ll a, ll b)
{
  return mod_mul(a, mod_pow(b, MOD - 2));
}

void make_fact(ll n)
{
  fac[0] = 1;
  REP(i, n)
  {
    fac[i + 1] = mod_mul(fac[i], i + 1);
  }
}

void make_invfact(ll n)
{
  invfac[n] = mod_pow(fac[n], MOD - 2);
  REP_R(i, n)
  {
    invfac[i] = mod_mul(invfac[i + 1], i + 1);
  }
}

void init_fact()
{
  make_fact(FACSIZE - 1);
  make_invfact(FACSIZE - 1);
}

ll combination(ll n, ll c)
{

  return mod_mul(fac[n], mod_mul(invfac[c], invfac[n - c]));
}
///////////////fact
/*
  init_fact();
  cout << fac[5] << endl;
  cout << invfac[5] << endl;
  cout << combination(8, 4) << endl;
  cout << combination(31, 8) << endl;
  //*/
///////////////////