stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

int n;

int main()

{

    cin>>n;

    bool f=false;

    for(int i=0;i<=25;i++){

        for(int j=0;j<=14;j++){

            if(4*i+7*j==n){

                f=true;

                break;

            }

        }

    }

    if(f)

        cout<<"Yes";

    else

        cout<<"No";

    return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

    int N,i,j;

	int sum=0;

	scanf("%d",&N);

    for(i=0;i<=25;i++)

	{

		for(j=0;j<=14;j++)

		{

            if(i*4+j*7==N)

			{

				sum=1;

			}

		}

	}

	if(sum==1)

	{

		printf("Yes\n");

	}

	if(sum==0)

	{

		printf("No\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<cstdio>

#include<iostream>

#include<cmath>

using namespace std;



int flag;



void dfs(int n){

     if (n < 0) return ;

     if(n == 0){

         flag = 1;

         //cout << "Yes"<<endl;

         return ;

     }

     dfs(n - 4);

     dfs(n - 7);

}



int main(){



   int n;

   while(cin >> n ){

        flag = 0;

        dfs(n);

        if(!flag) cout << "No" <<endl;

        else  cout << "Yes" <<endl;



   }

  return 0;

}


// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

    int N,i,j;

	int sum=0;

	scanf("%d",&N);

    for(i=0;i<=25;i++)

	{

		for(j=0;j<=14;j++)

		{

            if(i*4+j*7==N)

			{

				sum=1;

			}

		}

	}

	if(sum==1)

	{

		printf("Yes\n");

	}

	if(sum==0)

	{

		printf("No\n");

	}

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>



#define rep(i,n) for(long long int (i)=0;(i)<(int)(n);(i)++)

#define rrep(i,a,b) for(long long int i=(a);i<(b);i++)

#define rrrep(i,a,b) for(long long int i=(a);i>=(b);i--)

#define all(v) (v).begin(), (v).end()

#define pb(q) push_back(q)

#define Abs(a,b) max(a,b)-min(a,b)

#define YES(condition) if(condition){cout << "YES" << endl;}else{cout << "NO" << endl;}

#define Yes(condition) if(condition){cout << "Yes" << endl;}else{cout << "No" << endl;}

#define Cout(x) cout<<(x)<<endl

#define POSSIBLE(condition) if(condition){cout << "POSSIBLE" << endl;}else{cout << "IMPOSSIBLE" << endl;}

#define Possible(condition) if(condition){cout << "Possible" << endl;}else{cout << "Impossible" << endl;}

#define possible(condition) if(condition){cout << "possible" << endl;}else{cout << "impossible" << endl;}

#define Size(n) (n).size()



typedef long long ll;





using namespace std;



const int INF = 1e9,MOD = 1e9 + 7,ohara = 1e6;

const ll LINF = 1e18;





long long int n,cnt=0,ans=0,a,b,c,d,cmp,cmpp,m,h,w,x,y,sum=0,pos;

int dy[]={1,0,-1,0};

int dx[]={0,1,0,-1};

string alph("abcdefghijklmnopqrstuvwxyz"),s;

bool fl;

struct edge{int to,cost;};



//-------------------------↓↓↓↓↓↓------------------------



int main(void){

       cin.tie(0);

    ios::sync_with_stdio(false);



      

      cin>>n;

      rep(i,100){

          rep(j,100){

              pos=4*i+7*j;

              if(pos==n){

                  fl=true;

              }

          }

      }

      Yes(fl);

     



       

return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

int main(void){

    // Your code here!

    int N,ans=0;

    scanf("%d",&N);

    for(int i = 0;i <= 25;i++){

        for(int j = 0;j <= 15;j++){

            if(4 * i + 7 * j == N)ans = 1;

        }

    }

    if(ans == 1)printf("Yes\n");

    else printf("No\n");

    return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

using namespace std;



int main()

{

   int sx, sy, tx, ty;

    cin >> sx >> sy >> tx >> ty;

    int i;

    for (i=sy+1 ; i<=ty ; i++)

    {

        cout << "U";

    }

    for (i=sx+1 ; i<=tx ; i++)

    {

        cout << "R";

    }

    for (i=ty-1 ; i>=sy ; i--)

    {

        cout << "D";

    }

    for (i=tx-1 ; i>=sx ; i--)

    {

        cout << "L";

    }

    cout << "L";

    for (i=sy ; i<=ty ; i++)

    {

        cout << "U";

    }

    for (i=sx ; i<=tx ; i++)

    {

        cout << "R";

    }

    cout << "DR";

      

    for (i=ty ; i>=sy ; i--)

    {

        cout << "D";

    }

    for (i=tx ; i>=sx ; i--)

    {

        cout << "L";

    }

    cout << "U" ;

    return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;



int main()

{

	int a,b,x,y;

	while(~scanf("%d%d%d%d",&a,&b,&x,&y))

	{

		int n = x-a,m = y-b;

		for(int i = 0;i < m;i++)

			printf("U");

		for(int i = 0;i < n;i++)

			printf("R");

		for(int i = 0;i < m;i++)

			printf("D");

		for(int i = 0;i < n+1;i++)

			printf("L");

		for(int i = 0;i < m+1;i++)

			printf("U");

		for(int i = 0;i < n+1;i++)

			printf("R");

			printf("D");	

			printf("R");

		for(int i = 0;i < m+1;i++)

			printf("D");

		for(int i = 0;i < n+1;i++)

			printf("L");

		printf("U\n");

	}

return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

using namespace std;

int main()

{

	int x1,y1,x2,y2;

	cin>>x1>>y1>>x2>>y2;

	for(int i=1;i<=y2-y1;i++)

	cout<<"U";

	for(int i=1;i<=x2-x1;i++)

	cout<<"R";

	for(int i=1;i<=y2-y1;i++)

	cout<<"D";

	for(int i=1;i<=x2-x1+1;i++)

	cout<<"L";

	for(int i=1;i<=y2-y1+1;i++)

	cout<<"U";

	for(int i=1;i<=x2-x1+1;i++)

	cout<<"R";

	cout<<"D";

	cout<<"R";

	for(int i=1;i<=y2-y1+1;i++)

	cout<<"D";

	for(int i=1;i<=x2-x1+1;i++)

	cout<<"L";

	cout<<"U"<<endl;

	return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

#define INF 0x3f3f3f3f



char s[100];

typedef long long ll;



void step(int x, int y)

{

    for (int i = 0; i < y; ++i)

    {

        printf("U");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("R");

    }

    for (int i = 0; i < y; ++i)

    {

        printf("D");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("L");

    }

}



int main()

{

    //freopen("C:\\Users\\tiger\\Desktop\\a.txt","r",stdin);

	int x1,y1;

    int x2,y2;

    while ( ~scanf("%d %d %d %d", &x1, &y1, &x2, &y2) )

    {

        int x = x2 - x1;

        int y = y2 - y1;

        step(x,y);

        printf("L");

        for (int i = 0; i <= y; ++i)

        {

            printf("U");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("R");

        }

        printf("D");

        printf("R");

        for (int i = 0; i <= y; ++i)

        {

            printf("D");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("L");

        }

        printf("U\n");

    }

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
//============================================================================

// Name        : test.cpp

// Author      : Qihan

// Version     :

// Copyright   : Your copyright notice

// Description : Hello World in C++, Ansi-style

//============================================================================



#include <bits/stdc++.h>

#define pi acos(-1.0)

using namespace std;

typedef long long int LLI;

typedef pair<LLI,LLI> PII;

#define Lowbit(x) (x & (-x))

const int inf = 0x3f3f3f;

const int maxn = (500000 + 10);





int main() {

//    freopen("/home/qihan/Documents/in","r",stdin);

//    freopen("/home/qihan/Documents/out","w",stdout);

	int sx,sy,ex,ey;

	scanf("%d%d%d%d",&sx,&sy,&ex,&ey);

	int x = ex - sx,y = ey - sy;

	for(int i = 1;i <= y;i ++)  printf("U");

	for(int i = 1;i <= x;i ++)  printf("R");

	for(int i = 1;i <= y;i ++)  printf("D");

	for(int i = 1;i <= x;i ++)  printf("L");

	printf("L");

	for(int i = 1;i <= y + 1;i ++)  printf("U");

	for(int i = 1;i <= x + 1;i ++)  printf("R");

	printf("D");

	printf("R");

	for(int i = 1;i <= y + 1;i ++)  printf("D");

	for(int i = 1;i <= x + 1;i ++)  printf("L");

	printf("U\n");

	return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

#define INF 0x3f3f3f3f



char s[100];

typedef long long ll;



void step(int x, int y)

{

    for (int i = 0; i < y; ++i)

    {

        printf("U");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("R");

    }

    for (int i = 0; i < y; ++i)

    {

        printf("D");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("L");

    }

}



int main()

{

    //freopen("C:\\Users\\tiger\\Desktop\\a.txt","r",stdin);

	int x1,y1;

    int x2,y2;

    while ( ~scanf("%d %d %d %d", &x1, &y1, &x2, &y2) )

    {

        int x = x2 - x1;

        int y = y2 - y1;

        step(x,y);

        printf("L");

        for (int i = 0; i <= y; ++i)

        {

            printf("U");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("R");

        }

        printf("D");

        printf("R");

        for (int i = 0; i <= y; ++i)

        {

            printf("D");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("L");

        }

        printf("U\n");

    }

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int a[100000];

int main()

{

    int sx, sy, tx, ty;

    cin >> sx >> sy >> tx >> ty;

    for(int i=1;i<=ty-sy;i++)

    {

        cout<< "U";

    }

    for(int i=1;i<=tx-sx;i++)

    {

        cout<< "R";

    }

    for(int i=1;i<=ty-sy;i++)

    {

        cout<< "D";

    }

    for(int i=1;i<=tx-sx;i++)

    {

        cout<< "L";

    }

    cout<< "L";

    for(int i=0;i<=ty-sy;i++)

    {

        cout<< "U";

    }

    cout<< "R";

    for(int i=1;i<=tx-sx;i++)

    {

        cout<< "R";

    }

    cout<< "D";

    cout<< "R";

    for(int i=0;i<=ty-sy;i++)

    {

        cout<< "D";

    }

    cout<< "L";

    for(int i=1;i<=tx-sx;i++)

    {

        cout<< "L";

    }

    cout <<"U" <<endl;  

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

#define INF 0x3f3f3f3f



char s[100];

typedef long long ll;



void step(int x, int y)

{

    for (int i = 0; i < y; ++i)

    {

        printf("U");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("R");

    }

    for (int i = 0; i < y; ++i)

    {

        printf("D");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("L");

    }

}



int main()

{

    //freopen("C:\\Users\\tiger\\Desktop\\a.txt","r",stdin);

	int x1,y1;

    int x2,y2;

    while ( ~scanf("%d %d %d %d", &x1, &y1, &x2, &y2) )

    {

        int x = x2 - x1;

        int y = y2 - y1;

        step(x,y);

        printf("L");

        for (int i = 0; i <= y; ++i)

        {

            printf("U");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("R");

        }

        printf("D");

        printf("R");

        for (int i = 0; i <= y; ++i)

        {

            printf("D");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("L");

        }

        printf("U\n");

    }

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <cstring>

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <string>

#include <map>

#include <stack>

#include <queue>

#include <vector>

#define inf 0x3f3f3f3f

#define met(a,b) memset(a,b,sizeof a)

#define pb push_back

typedef long long ll;

using namespace std;

const int N = 20050;

const int M = 24005;

const int mod=1e9+7;

int n,m;

char str[N];

int main(){

    int tot=0;

    int sx,sy,ex,ey;

    scanf("%d%d%d%d",&sx,&sy,&ex,&ey);

    int x=ex-sx,y=ey-sy;

    for(int i=1;i<=y;i++)str[tot++]='U';

    for(int i=1;i<=x;i++)str[tot++]='R';

    for(int i=1;i<=y;i++)str[tot++]='D';

    for(int i=1;i<=x;i++)str[tot++]='L';

    str[tot++]='L';

    for(int i=1;i<=y+1;i++)str[tot++]='U';

    for(int i=1;i<=x+1;i++)str[tot++]='R';

    str[tot++]='D';

    str[tot++]='R';

    for(int i=1;i<=y+1;i++)str[tot++]='D';

    for(int i=1;i<=x+1;i++)str[tot++]='L';

    str[tot++]='U';

    str[tot]=0;

    printf("%s",str);

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

#define INF 0x3f3f3f3f



char s[100];

typedef long long ll;



void step(int x, int y)

{

    for (int i = 0; i < y; ++i)

    {

        printf("U");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("R");

    }

    for (int i = 0; i < y; ++i)

    {

        printf("D");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("L");

    }

}



int main()

{

    //freopen("C:\\Users\\tiger\\Desktop\\a.txt","r",stdin);

	int x1,y1;

    int x2,y2;

    while ( ~scanf("%d %d %d %d", &x1, &y1, &x2, &y2) )

    {

        int x = x2 - x1;

        int y = y2 - y1;

        step(x,y);

        printf("L");

        for (int i = 0; i <= y; ++i)

        {

            printf("U");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("R");

        }

        printf("D");

        printf("R");

        for (int i = 0; i <= y; ++i)

        {

            printf("D");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("L");

        }

        printf("U\n");

    }

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

int a[100000];

int main()

{

    int sx, sy, tx, ty;

    cin >> sx >> sy >> tx >> ty;

    for(int i=1;i<=ty-sy;i++)

    {

        cout<<"U";

    }

    for(int i=1;i<=tx-sx;i++)

    {

        cout<<"R";

    }

    for(int i=1;i<=ty-sy;i++)

    {

        cout<<"D";

    }

    for(int i=1;i<=tx-sx;i++)

    {

        cout<<"L";

    }

    cout<<"L";

    for(int i=0;i<=ty-sy;i++)

    {

        cout<<"U";

    }

    cout<<"R";

    for(int i=1;i<=tx-sx;i++)

    {

        cout<<"R";

    }

    cout<<"D";

    cout<<"R";

    for(int i=0;i<=ty-sy;i++)

    {

        cout<<"D";

    }

    cout<<"L";

    for(int i=1;i<=tx-sx;i++)

    {

        cout<<"L";

    }

    cout<<"U";

    return 0;

}

// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

#define INF 0x3f3f3f3f



char s[100];

typedef long long ll;



void step(int x, int y)

{

    for (int i = 0; i < y; ++i)

    {

        printf("U");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("R");

    }

    for (int i = 0; i < y; ++i)

    {

        printf("D");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("L");

    }

}



int main()

{

    //freopen("C:\\Users\\tiger\\Desktop\\a.txt","r",stdin);

	int x1,y1;

    int x2,y2;

    while ( ~scanf("%d %d %d %d", &x1, &y1, &x2, &y2) )

    {

        int x = x2 - x1;

        int y = y2 - y1;

        step(x,y);

        printf("L");

        for (int i = 0; i <= y; ++i)

        {

            printf("U");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("R");

        }

        printf("D");

        printf("R");

        for (int i = 0; i <= y; ++i)

        {

            printf("D");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("L");

        }

        printf("U\n");

    }

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define rep(i,a,n) for (int i=a;i<=n;i++)

#define per(i,a,n) for (int i=n;i>=a;i--)

#define pb push_back

#define mp make_pair

#define all(x) (x).begin(),(x).end()

#define fi first

#define se second

#define SZ(x) ((int)(x).size())

using namespace std;

typedef vector<int> vi;

typedef long long ll;

typedef pair<int,int> pii;

const ll mod=1000000007;

ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}

const int maxn=1e5+10;



int main()

{

    int sx,sy,tx,ty;

    scanf("%d%d%d%d",&tx,&ty,&sx,&sy);

    for(int i = ty; i < sy; ++i) {

        printf("U");

    }

    for(int i = tx; i < sx; ++i) {

        printf("R");

    }

     for(int i = ty; i < sy; ++i) {

        printf("D");

    }

    for(int i = tx; i < sx; ++i) {

        printf("L");

    }

    printf("L");

     for(int i = ty; i < sy+1; ++i) {

        printf("U");

    }

    for(int i = tx; i < sx+1; ++i) {

        printf("R");

    }

    printf("DR");

     for(int i = ty; i < sy+1; ++i) {

        printf("D");

    }

    for(int i = tx; i < sx+1; ++i) {

        printf("L");

    }

    printf("U\n");

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

#define INF 0x3f3f3f3f



char s[100];

typedef long long ll;



void step(int x, int y)

{

    for (int i = 0; i < y; ++i)

    {

        printf("U");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("R");

    }

    for (int i = 0; i < y; ++i)

    {

        printf("D");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("L");

    }

}



int main()

{

    //freopen("C:\\Users\\tiger\\Desktop\\a.txt","r",stdin);

	int x1,y1;

    int x2,y2;

    while ( ~scanf("%d %d %d %d", &x1, &y1, &x2, &y2) )

    {

        int x = x2 - x1;

        int y = y2 - y1;

        step(x,y);

        printf("L");

        for (int i = 0; i <= y; ++i)

        {

            printf("U");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("R");

        }

        printf("D");

        printf("R");

        for (int i = 0; i <= y; ++i)

        {

            printf("D");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("L");

        }

        printf("U\n");

    }

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

void print(char* c,int n)

{

    for(int i=0 ; i<n ; i++){

        cout << c;

    }

}

int main()

{

    int sx, sy, tx, ty;

    cin >> sx >> sy >> tx >> ty;

    print("U",ty-sy);

    print("R",tx-sx);

    print("D",ty-sy);

    print("L",tx-sx);

    print("L",1);

    print("U",ty-sy+1);

    print("R",tx-sx+1);

    print("D",1);

    print("R",1);

    print("D",ty-sy+1);

    print("L",tx-sx+1);

    print("U",1);

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

#define INF 0x3f3f3f3f



char s[100];

typedef long long ll;



void step(int x, int y)

{

    for (int i = 0; i < y; ++i)

    {

        printf("U");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("R");

    }

    for (int i = 0; i < y; ++i)

    {

        printf("D");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("L");

    }

}



int main()

{

    //freopen("C:\\Users\\tiger\\Desktop\\a.txt","r",stdin);

	int x1,y1;

    int x2,y2;

    while ( ~scanf("%d %d %d %d", &x1, &y1, &x2, &y2) )

    {

        int x = x2 - x1;

        int y = y2 - y1;

        step(x,y);

        printf("L");

        for (int i = 0; i <= y; ++i)

        {

            printf("U");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("R");

        }

        printf("D");

        printf("R");

        for (int i = 0; i <= y; ++i)

        {

            printf("D");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("L");

        }

        printf("U\n");

    }

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <stdio.h>

#include <cstring>

using namespace std;



int main()

{

    int a,b,x,y;

    while(scanf("%d %d %d %d",&a,&b,&x,&y)!=EOF)

    {

        for(int i=0;i<y-b;++i)

            cout<<'U';

        for(int i=0;i<x-a;++i)

            cout<<'R';

        for(int i=0;i<y-b;++i)

            cout<<'D';

        for(int i=0;i<x-a;++i)

            cout<<'L';

        cout<<'L';

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

            cout<<'U';

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

            cout<<'R';

            cout<<"DR";

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

            cout<<'D';

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

            cout<<'L';

        cout<<'U'<<endl;

    }

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

#define INF 0x3f3f3f3f



char s[100];

typedef long long ll;



void step(int x, int y)

{

    for (int i = 0; i < y; ++i)

    {

        printf("U");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("R");

    }

    for (int i = 0; i < y; ++i)

    {

        printf("D");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("L");

    }

}



int main()

{

    //freopen("C:\\Users\\tiger\\Desktop\\a.txt","r",stdin);

	int x1,y1;

    int x2,y2;

    while ( ~scanf("%d %d %d %d", &x1, &y1, &x2, &y2) )

    {

        int x = x2 - x1;

        int y = y2 - y1;

        step(x,y);

        printf("L");

        for (int i = 0; i <= y; ++i)

        {

            printf("U");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("R");

        }

        printf("D");

        printf("R");

        for (int i = 0; i <= y; ++i)

        {

            printf("D");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("L");

        }

        printf("U\n");

    }

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<cstdio>

#include<cstring>

#include<queue>

#include<algorithm>

#include<cmath>

#include<iostream>

using namespace std;

void print(int x,int y)

{

    int n=x,m=y;

    while(n--)

        putchar('U');

    while(m--)

        putchar('R');

    n=x,m=y;

    while(n--)

        putchar('D');

    while(m--)

        putchar('L');

    putchar('L');

    n=x,m=y;

    while(n--)

        putchar('U');

    putchar('U');

    while(m--)

        putchar('R');

    putchar('R');

    putchar('D');

    putchar('R');

    n=x,m=y;

    while(n--)

        putchar('D');

    putchar('D');

    while(m--)

        putchar('L');

    putchar('L');

    putchar('U');

putchar('\n');

}

int main()

{

    int sx,sy,ex,ey;

    while(scanf("%d%d%d%d",&sx,&sy,&ex,&ey)!=-1)

        print(ey-sy,ex-sx);

    return 0;

}


// Optimized solution to the same problem (faster):
#include <stdio.h>

#include <stdlib.h>

#include <string.h>

#include <math.h>

#define INF 0x3f3f3f3f



char s[100];

typedef long long ll;



void step(int x, int y)

{

    for (int i = 0; i < y; ++i)

    {

        printf("U");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("R");

    }

    for (int i = 0; i < y; ++i)

    {

        printf("D");

    }

    for (int i = 0; i < x; ++i)

    {

        printf("L");

    }

}



int main()

{

    //freopen("C:\\Users\\tiger\\Desktop\\a.txt","r",stdin);

	int x1,y1;

    int x2,y2;

    while ( ~scanf("%d %d %d %d", &x1, &y1, &x2, &y2) )

    {

        int x = x2 - x1;

        int y = y2 - y1;

        step(x,y);

        printf("L");

        for (int i = 0; i <= y; ++i)

        {

            printf("U");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("R");

        }

        printf("D");

        printf("R");

        for (int i = 0; i <= y; ++i)

        {

            printf("D");

        }

        for (int i = 0; i <= x; ++i)

        {

            printf("L");

        }

        printf("U\n");

    }

	return 0;

}

// Fastest known solution for this problem: