Problem:
Score : 300 points

Problem Statement

Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up.

Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1.

Here, both the x- and y-coordinates before and after each movement must be integers.

He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy).

Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty).

Under this condition, find a shortest path for him.

Constraints

-1000 ≤ sx < tx ≤ 1000

-1000 ≤ sy < ty ≤ 1000

sx,sy,tx and ty are integers.

Input

The input is given from Standard Input in the following format:

sx sy tx ty

Output

Print a string S that represents a shortest path for Dolphin.

The i-th character in S should correspond to his i-th movement.

The directions of the movements should be indicated by the following characters:

U: Up

D: Down

L: Left

R: Right

If there exist multiple shortest paths under the condition, print any of them.

Sample Input 1

0 0 1 2

Sample Output 1

UURDDLLUUURRDRDDDLLU

One possible shortest path is:

Going from (sx,sy) to (tx,ty) for the first time: (0,0) → (0,1) → (0,2) → (1,2)

Going from (tx,ty) to (sx,sy) for the first time: (1,2) → (1,1) → (1,0) → (0,0)

Going from (sx,sy) to (tx,ty) for the second time: (0,0) → (-1,0) → (-1,1) → (-1,2) → (-1,3) → (0,3) → (1,3) → (1,2)

Going from (tx,ty) to (sx,sy) for the second time: (1,2) → (2,2) → (2,1) → (2,0) → (2,-1) → (1,-1) → (0,-1) → (0,0)

Sample Input 2

-2 -2 1 1

Sample Output 2

UURRURRDDDLLDLLULUUURRURRDDDLLDL

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

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

#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 
  
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 
 
#define FS              ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll              long long int
#define ld              long double
#define pb              push_back
#define bp              __builtin_popcount
#define sz              size()
#define ff              first
#define ss              second
#define vll             vector<ll>
#define vbool           vector<bool>
#define vpll            vector<pair<ll,ll>>
#define pll             pair<ll,ll>
#define vllv            vector<vector<ll>>
#define setpri(x)       cout<<setprecision(x)<<fixed;
#define all(v)          v.begin(),v.end()
#define allr(v)         v.rbegin(),v.rend()
#define yesr {cout<<"YES"<<endl;return;}
#define nor {cout<<"NO"<<endl;return;}
// getline (std::cin,name);
ll MOD=1e9+7;
ll ceil1(ll n,ll x){return (n-1)/x+(n>0);}

ll gcd(ll a,ll b){return __gcd(a,b);}
ll lcm(ll a,ll b){return (max(a,b)/gcd(a,b))*min(a,b);}

ll pow1(ll n,ll m ,ll mod=MOD );
ll pow2(ll n,ll k);

ll modinv(ll n,ll mod=MOD){  return pow1(n,mod-2,mod);}

bool func(pair<ll,ll> &a,pair<ll,ll> &b ){
        if(a.ff != b.ff)return a.ff < b.ff;
        return a.ss > b.ss;
}

ll const N=(ll)3e5+11;
ll const LG=(ll)log2(N)+1;


void solve(){
        ll i,j,k,l,n,m,x,y,z,r;
                ll x1,x2,y1,y2;
                cin>>x1>>y1>>x2>>y2;
              string s1,s2,s3,s4;
              s1="";
              s2="";
              s3="";
              s4="";
              for(i=y1;i<y2;i++){
                      s1+='U';
                      s2+='D';
              }
              for(i=x1;i<x2;i++){
                      s3+='R';
                      s4+='L';
              }
              cout<<s1<<s3<<s2<<s4<<"L"<<s1<<"U"<<s3<<"RDLD"<<s2<<s4<<"LU";
              
              
        
        
}       
int main(){
        FS;
        
        ll i,j,k,n,m,x,y,z,q;
        q=1;
        // cin>>q;
        for(ll tc=1;tc<=q;tc++){
                // cout<<"Case #"<<tc<<": ";
                solve();
        }
        return 0;
}


ll pow1(ll n,ll m,ll mod){
        if(m==0)return 1;
        
        if(m%2==0)return pow1((n*n)%mod,m/2,mod);
    	
    	return (pow1((n*n)%mod,m/2,mod)*n)%mod;
}

ll pow2(ll n,ll k){
        ll ans=1;
            while(k>0){
                if(k%2==1)ans=ans*n;
                n=n*n;
                k/=2;
            }
            return ans;
}