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≤10¹⁶+1000

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Input

Input is given from Standard Input in the following format:

Na1 a2 ... aN

Output

Print the number of times the operation will be performed.

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Sample Input 1

Copy

43 3 3 3

Sample Output 1

Copy

0 ABC78 D题的逆向，可以直接模拟。

1 #include 
      
        2 #define ll long long 3 using namespace std; 4 ll a[55]; 5 int main() { 6     ios::sync_with_stdio(false); 7     int n; 8     cin >> n; 9     for(int i = 1; i <= n; i ++) {10         cin >> a[i];11     }12     ll k = 0;13     while(1) {14         ll MAX = -1,id;15         for(int i = 1; i <= n; i ++) {16             if(MAX < a[i]){17                 MAX = a[i];18                 id = i;19             }20         }21         if(MAX < n)break;22         for(int i = 1; i <= n; i ++) {23             if(i == id) a[i] %= n;24             else a[i] += MAX/n;25         }26         k += MAX / n;27     }28     printf("%lld\n",k);29     return 0;30 }