Problem: You are standing in front of the room with great treasures. The only thing stopping you is the door with a push-button combination lock. This lock has d buttons with digits from 0 to d - 1. Whenever you press a button, it stays pushed down. You can not pop back up just one button, but there is a "RESET" button — pressing it pops up all other buttons. Initially, no buttons are pushed down.

The door instantly opens when some specific set of digits is pushed down. Sadly, you don't know the password for it. Having read the documentation for this specific lock, you found out that there are n possible passwords for this particular lock. 

Find the shortest sequence of button presses, such that all possible passwords appear at least once during its execution. Any shortest correct sequence of button presses will be accepted.

Input

The first line contains two integers d and n (1 ≤ d ≤ 10; 1 ≤ n ≤ 2^d - 1). Next n lines describe possible passwords. Each line contains a string s_i of d zeros and ones: for all j from 1 to d the j-th character is 1 iff the button with the digit j - 1 must be pushed down.

All strings s_i are different, and each string contains at least one 1.

Output

On the first line, print the number k — the minimum number of button presses. On the second line, print k tokens, describing the sequence. Whenever you press a button with a digit, print that digit. Whenever you press "RESET", print "R".

Examples

Input


2 2
10
11


Output


2
0 1 


Input


3 4
001
111
101
011


Output


6
2 0 R 1 2 0 

Note

In the second example, the sequence 1 2 R 2 0 1 is also possible.

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

const int N = 1070;
const int inf = 10007;
int visx[N], visy[N], lx[N], ly[N], G[N][N], nx, ny;
int fa[N<<1], match[N], slack[N];
int findpath(int x) {
    int tempDelta;
    visx[x]=true;
    for(int y=1;y<=ny;y++){
        if(visy[y])continue;
        tempDelta =lx[x]+ly[y]-G[x][y];
        if(tempDelta == 0){
            visy[y] = true;
            fa[y+nx]=x;
            if(!match[y]) return y+nx;
            fa[match[y]]=y+nx;//记录交替树的父亲信息（为了区别X，Y集合，Y的点都映射成n+y）
            int res=findpath(match[y]);
            if(res>0)return res;//返回增广路的末端叶子节点
        }
        else if(slack[x] > tempDelta)//统计以x为准的slack值。
            slack[x] = tempDelta;
    }
    return -1;
}
void KM() {
    for(int x = 1; x <= nx; ++x) lx[x] = inf;
    for(int x = 1 ; x <= nx ; ++x) {
       for(int i = 1 ; i <= nx ; ++i) slack[i] = inf;
       for(int i=1;i<=nx+ny;i++)fa[i]=-1;
        memset(visx,false,sizeof(visx));
        memset(visy,false,sizeof(visy));//换到外面，可以保留原树
        int fir=1, leaf=-1;
        while(true){
            if(fir==1){
                leaf=findpath(x);
                fir=0;
            } else{
                for(int i=1;i<=nx;i++){
                    if(slack[i]==0){//只接着搜有新边加入的X点
                        slack[i] = inf;//slack要重新清空，方以后接着用
                        leaf=findpath(i);
                        if(leaf>0)break;
                    }
                }
            }
            if(leaf > 0) {
                int p=leaf;
                while(p>0){
                    match[p-nx]=fa[p];
                    p=fa[fa[p]];//顺着记录一路找找上去
                }
                break;
            } else {
                int delta = inf;
                for(int i = 1 ; i <= nx ; ++i)
                    if(visx[i] && delta > slack[i])
                        delta = slack[i];
                for(int i = 1 ; i <= nx ; ++i)
                    if(visx[i]) {lx[i] -= delta;slack[i]-=delta;}//X点的slack要响应改变，slack变0说明有新边加入
                for(int j = 1 ; j <= ny ; ++j){
                    if(visy[j])
                        ly[j] += delta;
                }
            }
        }
    }
}

int popcnt(int x) {
    int cnt = 0;
    for(int i = 0; i <= 10; ++i) {
        if((1<<i)&x) ++cnt;
    }
    return cnt;
}

int tag[N], nxt[N], pre[N];
char op[20], oo[10005];
int main() {
    int d, m, n;
    cin >> d >> m;
    n = (1 << d) - 1;
    nx = ny = n;
    for(int i = 1; i <= m; ++i) {
        scanf("%s", op);
        int val = 0;
        for(int j = d-1; j>=0; --j) {
            val = 2 * val + (op[j] - '0');
        }
        tag[val] = 1;
    } 
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= n; ++j) {
            if(i == j) continue;
            if(tag[i] && tag[j] && (i | j) == i) {
                G[j][i] = popcnt(j) + 1;
            }
        }
    }
    KM();
    int ans = m-1;
    for(int x = 1; x <= n; ++x) {
        if(tag[x]) ans += popcnt(x); 
    }
    for(int y = 1, x; y <= ny; ++y) {
        if(match[y]) ans -= G[match[y]][y];
        x = match[y];
        if(tag[x] && tag[y] && G[x][y]) {
            pre[y] = x;
            nxt[x] = y;
        }
    }
    printf("%d\n", ans);
    int tot = 0;
    for(int x = 1; x <= n; ++x) {
        if(tag[x] && !pre[x]) {
            int cur = 0;
            for(int y = x; y; y = nxt[y]) {
                for(int i = 0; i <= 10; ++i) {
                    if(((1<<i)&y) && !((1<<i)&cur)) oo[++tot] = i + '0';
                }
                cur |= y;
            }
            oo[++tot] = 'R';
        }
    }
    for(int i = 1; i < tot; ++i) {
        printf("%c ", oo[i]);
    }
    printf("\n");
}


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