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;
 
#define SPEED ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define fileio freopen("in.in", "r", stdin),freopen("out.out", "w", stdout);
#define ll long long
#define int long long
#define FF first
#define SS second
#define mp make_pair
#define pb push_back
#define pii pair<ll,ll>
#define pll pair<long long,long long>
#define sd(x) scanf("%d",&x)
#define slld(x) scanf("%lld",&x)
#define pd(x) prllf("%d\n",x)
#define plld(x) prllf("%lld\n",x)
#define pss prllf
#define MOD 1000000007
#define eps 0.00001
#define endl '\n'
#define debug(n1) cout<<n1<<endl
ll n,k,s[1005],e[1005],c[1005];
bool y[1005];
const ll INF = 1e17+5;
int mm[100005];
vector<int>temp[100005];

struct MinimumCostMaximumFlow {
  typedef ll Index; typedef ll Flow; typedef ll Cost;
  static const Flow InfCapacity = INF;
  struct Edge {
    Index to; Index rev;
    Flow capacity; Cost cost;
  };
  vector<vector<Edge> > g;
  void init(Index n) { g.assign(n, vector<Edge>()); }
  void addEdge(bool inv,Index i, Index j, Flow capacity = InfCapacity, Cost cost = Cost() ) {
    Edge e, f; e.to = j, f.to = i; e.capacity = capacity, f.capacity = 0; e.cost = cost, f.cost = -cost;
    if(inv)
    {
        temp[i].push_back(mm[i]);
    }
    mm[i]++;
    mm[j]++;

    g[i].push_back(e); g[j].push_back(f);
    g[i].back().rev = (Index)g[j].size() - 1; g[j].back().rev = (Index)g[i].size() - 1;
  }
  void addB(Index i, Index j, Flow capacity = InfCapacity, Cost cost = Cost()) {
    addEdge(0,i, j, capacity, cost);
    addEdge(0,j, i, capacity, cost);
  }

  pair<Cost, Flow> minimumCostMaximumFlow(Index s, Index t, Flow f = InfCapacity, bool useSPFA = false) {
    ll n = g.size();
    vector<Cost> dist(n); vector<Index> prev(n); vector<Index> prevEdge(n);
    pair<Cost, Flow> total = make_pair(0, 0);
    vector<Cost> potential(n);
    while(f > 0) {
      fill(dist.begin(), dist.end(), INF);
      if(useSPFA || total.second == 0) {
        deque<Index> q;
        q.push_back(s); dist[s] = 0; vector<bool> inqueue(n);
        while(!q.empty()) {
          Index i = q.front(); q.pop_front(); inqueue[i] = false;
          for(Index ei = 0; ei < (Index)g[i].size(); ei ++) {
            const Edge &e = g[i][ei]; Index j = e.to; Cost d = dist[i] + e.cost;
            if(e.capacity > 0 && d < dist[j]) {
              if(!inqueue[j]) {
                inqueue[j] = true;
                q.push_back(j);
              }
              dist[j] = d; prev[j] = i; prevEdge[j] = ei;
            }
          }
        }
      } else {
        vector<bool> vis(n);
        priority_queue<pair<Cost, Index> > q;
        q.push(make_pair(-0, s)); dist[s] = 0;
        while(!q.empty()) {
          Index i = q.top().second; q.pop();
          if(vis[i]) continue;
          vis[i] = true;
          for(Index ei = 0; ei < (Index)g[i].size(); ei ++) {
            const Edge &e = g[i][ei];
            if(e.capacity <= 0) continue;
            Index j = e.to; Cost d = dist[i] + e.cost + potential[i] - potential[j];
            if(dist[j] > d) {
              dist[j] = d; prev[j] = i; prevEdge[j] = ei;
              q.push(make_pair(-d, j));
            }
          }
        }
      }
      if(dist[t] == INF) break;
      if(!useSPFA) for(Index i = 0; i < n; i ++) potential[i] += dist[i];
 
      Flow d = f; Cost distt = 0;
      for(Index v = t; v != s; ) {
        Index u = prev[v]; const Edge &e = g[u][prevEdge[v]];
        d = min(d, e.capacity); distt += e.cost; v = u;
      }
      f -= d; total.first += d * distt; total.second += d;
      for(Index v = t; v != s; v = prev[v]) {
        Edge &e = g[prev[v]][prevEdge[v]];
        e.capacity -= d; g[e.to][e.rev].capacity += d;
      }
    }
    return total;
  }
};
 

int find_num(string ss)
{
    int ans=0;
    int tt=1;
    reverse(ss.begin(),ss.end());

    for(auto i:ss)
    {
        ans+=(i-'0')*tt;
        tt*=2;
    }
    return ans;
}



vector<vector<int> >adj;
int sz[1000005];
int vis[1000005];
int a[100005];
vector<string>fi;
int d;

void print(int yy)
{
    for(int i=0;i<10;i++)
    {
        if(((1<<i)&yy))
        {
            fi.push_back(to_string(d-i-1));
        }
    }
}

void dfs(int ver,int x)
{
    vis[ver]=1;
    print(a[ver]^x);
    int cnt=0;


    for(auto i:adj[ver])
    {
        if(vis[i]==0)
        {
            if(cnt)
            {
                fi.push_back("R");
                print(a[ver]);
                dfs(i,a[ver]);
            }
            else
            {
                dfs(i,a[ver]);
            }
            cnt++;
        }
    }
}

signed main()
{
 
ios_base::sync_with_stdio(false); 
cin.tie(NULL);  
cout.tie(0);   
 
#ifndef ONLINE_JUDGE
    if(fopen("INPUT.txt","r"))
    {
        freopen("INPUT.txt","r",stdin);
        freopen("OUTPUT.txt","w",stdout);
    }

#endif          

    int n;
    cin>>d>>n;
    string se[n];
    for(int i=0;i<n;i++)
    {
        cin>>se[i];
        a[i+1]=find_num(se[i]);
    }

    adj.resize(n+1);

    MinimumCostMaximumFlow kad;
    int source=0;
    int sink=2*n+5;

    kad.init(2*n+10);

    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(i!=j)
            {
                if((a[i]&a[j])==a[j])
                {
                    kad.addEdge(1,n+j,i,1,-(__builtin_popcount(a[j])+1));
                }
            }
        }

        kad.addEdge(0,i,n+i,1,0);
        kad.addEdge(0,source,i,1,0);
        kad.addEdge(0,n+i,sink,1,-(__builtin_popcount(a[i])+1));
    }

    kad.addEdge(0,source,sink,n,0);

    pair<int,int> ans=kad.minimumCostMaximumFlow(source,sink,n,0);

    for(int i=n+1;i<=2*n;i++)
    {
        for(auto j:temp[i])
        {
            if(kad.g[i][j].capacity==0)
            {
                adj[i-n].push_back(kad.g[i][j].to);
            }
        }
    }

    set<pair<int,int> >ss;
    int cc=0;
    for(int i=1;i<=n;i++)
    {
        ss.insert({a[i],i});
    }

    while(!ss.empty())
    {
        auto j=ss.begin();
        if(vis[j->second])
        {
            ss.erase(j);
            continue;
        }
        if(cc)
            fi.push_back("R");
        cc++;
        dfs(j->second,0);
    }

    cout<<fi.size()<<'\n';
    for(auto i:fi)
        cout<<i<<" ";



}