//Dinic+枚举字典序最小的最小割点集

//Time:1032Ms   Memory:1492K

#include<iostream>

#include<cstring>

#include<cstdio>

#include<algorithm>

#include<queue>

using namespace std;

#define MAXN 205

#define INF 0x3f3f3f3f

int N, S, T;

int s,t;

int sres[2*MAXN][2*MAXN];   //source-res

int res[2*MAXN][2*MAXN];

int d[2*MAXN];

int cut[MAXN];  //最小割点集

bool v[2*MAXN];

bool bfs()

{

    memset(d, -1, sizeof(d));

    queue<int> q;

    q.push(s);  d[s] = 0;

    while(!q.empty() && d[t] == -1){

        int cur = q.front();    q.pop();

        for(int i = 1; i <= t; i++)

        {

            if(d[i] == -1 && res[cur][i])

            {

                d[i] = d[cur] + 1;

                q.push(i);

            }

        }

    }

    return d[t] != -1;

}

int dfs(int x, int sum)

{

    if(x == t || sum == 0)  return sum;

    int src = sum;

    for(int i = 1; i <= t; i++)

    {

        if(d[i] == d[x] + 1 && res[x][i])

        {

            int tmp = dfs(i, min(sum, res[x][i]));

            res[x][i] -= tmp;

            res[i][x] += tmp;

            sum -= tmp;

        }

    }

    return src - sum;

}

int Dinic()

{

    memcpy(res, sres, sizeof(sres));

    int maxFlow = 0;

    while(bfs())

        maxFlow += dfs(s,INF);

    return maxFlow;

}

int main()

{

    //freopen("in.txt", "r", stdin);

    while(~scanf("%d%d%d", &N,&S,&T))

    {

        memset(sres, 0 ,sizeof(sres));

        s = 0;  t = 2*N+1;

        sres[0][S] = sres[T+N][t] = INF;

        for(int i = 1; i <= N; i++)

        {

            sres[i][i + N] = 1;

            for(int j = 1; j <= N; j++)

            {

                int num;

                scanf("%d", &num);

                if(num && i != j) sres[i+N][j] = INF;

            }

        }

        sres[S][S+N] = sres[T][T+N] = INF;

        int ans = Dinic();

        if(ans == INF){ //不可分开

            printf("NO ANSWER!\n");

            continue;

        }

        else {

            printf("%d\n", ans);

            if(ans == 0) continue;  //已经分开

        }

        //枚举最小割点集

        int len = 0, tmp = ans;

        for(int i = 1; i <= N && tmp; i++)

        {

            if(i == S || i == T)   continue;

            if(res[i][i+N])    continue;

            sres[i][i+N] = 0;

            int k = Dinic();

            if(k != tmp){

                tmp = k;

                cut[len++] = i;

            }

            else sres[i][i+N] = 1;

        }

        for(int i = 0; i < ans - 1; i++)

            printf("%d ", cut[i]);

        printf("%d\n", cut[ans-1]);

    }

    return 0;

}