Problem: A linear congruential generator produces a series R(⋅) of pseudo-random numbers by the following for- mulas:

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where S, A, C, and M are all parameters. In this problem, 0 ≤ S, A, C ≤ 15 and M = 256.

Now suppose we have some input string I(⋅), where each character in the string is an integer between 0 and (M - 1). Then, using the pseudo-random number series R(⋅), we obtain another string O(⋅) as the output by the following formula:

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Your task is to write a program that shows the parameters S, A, and C such that the information entropy of the output string O(⋅) is minimized. Here, the information entropy H is given by the following formula:

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where N is the length of the string and #(x) is the number of occurences of the alphabet x.



Input

The input consists of multiple datasets. Each dataset has the following format:

N
I(1) I(2) ... I(N)


N does not exceed 256.

The last dataset is followed by a line containing one zero. This line is not a part of any dataset and should not be processed.

Output

For each dataset, print in a line the values of the three parameters S, A, and C separated by a single space. If more than one solution gives the same minimum entropy, choose the solution with the smallest S, A, and then C.

Example

Input

5
5 4 3 2 1
5
7 7 7 7 7
10
186 8 42 24 154 40 10 56 122 72
0


Output

0 1 1
0 0 0
8 7 14

Solution (in CPP):

Solution:
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;

const int M = 256;
const double epsilon = 1e-9;

int main() {
  int N;
  while (cin >> N && N) {
    vector<int> L;
    for (auto i=0; i<N; i++) {
      int l;
      cin >> l;
      L.push_back(l);
    }
    double minH = 1e9;
    int minS, minA, minC;
    for (auto S=0; S<16; S++) {
      for (auto A=0; A<16; A++) {
	for (auto C=0; C<16; C++) {
	  int R[300];
	  R[0] = S;
	  for (auto i=1; i<=N; i++) {
	    R[i] = (A*R[i-1]+C)%M;
	    // cerr << R[i] << endl;
	  }
	  int O[300];
	  int App[300];
	  fill(App, App+300, 0);
	  for (auto i=0; i<N; i++) {
	    O[i] = (R[i+1] + L[i])%M;
	    // if (S == 1 && A == 5 && C == 11) cerr << O[i] << endl;
	    App[O[i]]++;
	  }
	  double H = 0;
	  for (auto i=0; i<300; i++) {
	    double temp = (double)App[i]/N;
	    // cerr << "temp = " << temp;
	    if (temp != 0) H -= temp * log(temp);
	  }
	  // cerr << H << endl;
	  if (H + epsilon < minH) { 
	    minH = H;
	    minS = S;
	    minA = A;
	    minC = C;
	    // cerr << "S = " << minS << ", A = " << minA 
	    //  << ", C = " << minC << ", H = " << H << endl;
	  }
	}  
      }
    }
    cout << minS << " " << minA << " " << minC << endl;
  }
}