Problem: In 2012, human beings have been exposed to fierce onslaught of unidentified mysterious extra-terrestrial creatures. We have exhaused because of the long war and can't regist against them any longer. Only you, an excellent wizard, can save us. Yes, it's time to stand up!

The enemies are dispatched to the earth with being aligned like an n * n square. Appearently some of them have already lost their fighting capabilities for some unexpected reason. You have invented a highest grade magic spell 'MYON' to defeat them all. An attack of this magic covers any rectangles (which is parallel to axis). Once you cast a spell "myon," then all the enemies which the magic covers will lose their fighting capabilities because of tremendous power of the magic. However, the magic seems to activate enemies' self-repairing circuit. Therefore if any enemy which have already lost its fighting capability is exposed to the magic, the enemy repossesses its fighting capability. You will win the war when all the enemies will lose their fighting capabilities.

Let me show an example. An array of enemies below have dispatched:


1 1 1 1 1
1 0 0 0 1
1 0 1 0 1
1 0 0 0 1
1 1 1 1 1


Here, '0' means an enemy that doesn't possess fighting capability, and '1' means an enemy that possesses. First, you cast a spell "myon" with covering all the enemies, which results in;


0 0 0 0 0
0 1 1 1 0
0 1 0 1 0
0 1 1 1 0
0 0 0 0 0


Next, cast once again with covering central 3 * 3 enemies.


0 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 0 0
0 0 0 0 0


Now you can defeat them all by casting a spell covers remaining one. Therefore you can cast "myonmyonmyon," which is the shortest spell for this case.

You are given the information of the array. Please write a program that generates the shortest spell that can defeat them all.

Constraints

* 1 ≤ n ≤ 5

Input

The first line of each test case has an integer n. Following n lines are information of an array of enemies. The format is same as described in the problem statement.

Input terminates when n = 0. You may assume that one or more enemy have fighting capability.

Output

For each dataset, output the shortest spell.

Example

Input

5
1 1 1 1 1
1 0 0 0 1
1 0 1 0 1
1 0 0 0 1
1 1 1 1 1
3
1 1 1
1 1 1
1 1 1
5
1 1 1 1 0
1 1 1 1 0
1 0 0 0 1
0 1 1 1 1
0 1 1 1 1
0


Output

myonmyonmyon
myon
myonmyon

Solution (in CPP):

Solution:
//Name: Mysterious Onslaught
//Level: 4
//Category: DP,ビット演算
//Note:

/*
 * 状態数は全部で 2^25 だが，これはちょうどメモリ制限の値と等しいため
 * 愚直な探索では解を求められない．
 * この手の問題だと上の行の状態を覚えておいて下の行に伝搬させるのが常套手段だが，
 * この問題でもその手法が使える．
 *
 * この問題で上の行から下の行へ伝搬する状態は，上の行でどのように矩形が置かれていたかである．
 * たとえばある行 r で 1〜3 列目を埋めるように矩形が置かれていた場合，その下の行 r+1 でも
 * 同じように 1〜3 列目を埋めるようにすると，これらはまとめて1個の矩形で表現できるため
 * コストが下がる．
 *
 * ある行について，列 x から右に矩形を伸ばしたとき，その幅は N-x 通り存在する．
 * これを各列について合計すると N=5 のときでも 15 通りであるから，1行を覆う矩形の取り方は
 * 高々 2^15 通り．
 * このうちその行を全滅させる置き方は更に少ないと予測できるので，そのような置き方だけを覚えて
 * 直前の状態として使うと計算量が抑えられる．
 * 実際には N=5 の時で 1024 通りになるので，1行の更新は 2^25 ループ程度で完了できる．
 *
 * 以下のコードではビット演算を使って，矩形の正当性(ちゃんと全てのビットを0にできるか)の判定を高速化している．
 * オーダーは O(2^(N^2)) ．
 */
#include <iostream>
#include <vector>

using namespace std;

const int INF = 30;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    while(true) {
        int N;
        cin >> N;
        if(!N) break;
        vector<int> field(N);
        for(int i = 0; i < N; ++i) {
            int pat = 0;
            for(int j = 0; j < N; ++j) {
                int v;
                cin >> v;
                pat <<= 1;
                pat |= v;
            }
            field[i] = pat;
        }
        // ビット演算で扱いやすくするため，右端を座標 0 として扱う．
        vector<int> myons; // 行にかかるmyon
        for(int c = 0; c < N; ++c) { // 右端の位置
            for(int w = 1; w <= N-c; ++w) {
                int mask = ((1<<w)-1) << c;
                myons.push_back(mask);
                //cout << mask << endl;
            }
        }
        const int M = myons.size();
        // 0: prev
        // 1: cur
        vector<int> valid_pats[2];
        vector<int> dp[2];
        dp[0].resize(1<<M, INF);
        dp[1].resize(1<<M, INF);
        for(int r = 0; r < N; ++r) {
            valid_pats[1].clear();
            fill(dp[1].begin(), dp[1].end(), INF);
            for(int pat = 0; pat < (1<<M); ++pat) {
                int mask = 0;
                for(int i = 0, ord = 1; i < M; ++i, ord <<= 1) {
                    if(pat & ord) mask ^= myons[i];
                }
                if(mask == field[r]) {
                    // Valid pattern
                    int cost = INF;
                    const int bits = __builtin_popcount(pat);
                    if(r != 0) {
                        // コストは
                        //   直前で有効な状態のコスト + 現在のパターンのコスト - 直前と合成する矩形のコスト
                        for(vector<int>::iterator it = valid_pats[0].begin(); it != valid_pats[0].end(); ++it) {
                            cost = min(cost, dp[0][*it] + bits - __builtin_popcount(pat & *it));
                        }
                    } else {
                        cost = bits;
                    }
                    //cout << r << ' ' << pat << ' ' << cost << endl;
                    dp[1][pat] = cost;
                    valid_pats[1].push_back(pat);
                }
            }
            valid_pats[0].swap(valid_pats[1]);
            dp[0].swap(dp[1]);
        }
        int ans = INF;
        for(vector<int>::iterator it = valid_pats[0].begin(); it != valid_pats[0].end(); ++it) {
            ans = min(ans, dp[0][*it]);
        }
        for(int i = 0; i < ans; ++i) {
            cout << "myon";
        }
        cout << endl;
    }
    return 0;
}