// Red Knight's Shortest Path
#include <bits/stdc++.h>
using namespace std;
#define GET_MACRO(_1, _2, _3, NAME, ...) NAME
#define _repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define _rep(i,n) _repl(i,0,n)
#define rep(...) GET_MACRO(__VA_ARGS__, _repl, _rep)(__VA_ARGS__)
#define mp(a,b) make_pair((a),(b))
#define pb(a) push_back((a))
#define all(x) (x).begin(),(x).end()
#define uniq(x) sort(all(x)),(x).erase(unique(all(x)),end(x))
#define fi first
#define se second
#define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
void _dbg(string){cout<<endl;}
template<class H,class... T> void _dbg(string s,H h,T... t){int l=s.find(',');cout<<s.substr(0,l)<<" = "<<h<<", ";_dbg(s.substr(l+1),t...);}
template<class T,class U> ostream& operator<<(ostream &o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream &o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

#define INF 1120000000
#define long long long // for codeforces

int dis[205][205];
int pre[205][205];

int main(){
  int n;
  int sx,sy,tx,ty;
  cin>>n>>sx>>sy>>tx>>ty;

  int dx[] = {-2,-2,0,2,2,0}, dy[] = {-1,1,2,1,-1,-2};
  fill(dis[0], dis[n], INF);
  fill(pre[0], pre[n], -1);

  dis[tx][ty] = 0;
  using P = pair<int,int>;
  priority_queue<pair<int, P>> pq;
  pq.push(mp(0, mp(tx,ty)));
  while(!pq.empty()){
    auto pp = pq.top(); pq.pop();
    int x = pp.se.fi, y = pp.se.se;
    int d = -pp.fi;
    if(d > dis[x][y]) continue;
    rep(k,6){
      int nx = x - dx[k], ny = y - dy[k];
      if(nx<0 || ny<0 || nx>=n || ny>=n) continue;
      if(d+1 < dis[nx][ny] || (d+1 == dis[nx][ny] && pre[nx][ny] > k)){
        dis[nx][ny] = d+1;
        pq.push(mp(-d-1, mp(nx,ny)));
        pre[nx][ny] = k;
      }
    }
  }

  if(dis[sx][sy] >= INF){ cout << "Impossible" << endl; return 0; }

  vector<int> res;
  while(pre[sx][sy] != -1){
    int p = pre[sx][sy];
    res.pb(p);
    sx += dx[p];
    sy += dy[p];
  }

  string xx [] = {"UL", "UR", "R", "LR", "LL", "L"};

  int m = res.size();
  cout << m << endl;
  rep(i,m) cout << xx[res[i]] << " \n"[i==m-1];

  return 0;
}