Problem: The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting his time, he wants to get from his current position s to square t in the least number of moves. Help him to do this.

<image>

In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).

Input

The first line contains the chessboard coordinates of square s, the second line — of square t.

Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8.

Output

In the first line print n — minimum number of the king's moves. Then in n lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD. 

L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them. 

Examples

Input

a8
h1


Output

7
RD
RD
RD
RD
RD
RD
RD

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int xx[4] = {1, -1, 0, 0};
int yy[4] = {0, 0, 1, -1};
const double eps = 1e-9;
const int maxn = 2e6 + 5000;
const long long mod = 1e9 + 7;
const int INF(0x3f3f3f3f);
const long long INFL(0x3f3f3f3f3f3f3f3fll);
inline int sign(double a) { return a < -eps ? -1 : a > eps; }
long long mul(long long a, long long b, long long c) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a, res %= c;
    a *= a, a %= c, b >>= 1;
  }
  return res;
}
long long phi(long long x) {
  long long res = x;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) res = res / i * (i - 1);
    while (x % i == 0) x /= i;
  }
  if (x > 1) res = res / x * (x - 1);
  return res;
}
template <typename A, typename B>
inline bool chmin(A& a, B b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename A, typename B>
inline bool chmax(A& a, B b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename A, typename B>
inline long long add(A x, B y) {
  if (x + y < 0) return x + y + mod;
  return x + y >= mod ? x + y - mod : x + y;
}
template <typename A, typename B>
inline void add2(A& x, B y) {
  if (x + y < 0)
    x = x + y + mod;
  else
    x = (x + y >= mod ? x + y - mod : x + y);
}
template <typename A, typename B>
inline long long mul1(A x, B y) {
  return 1ll * x * y % mod;
}
template <typename A, typename B>
inline void mul2(A& x, B y) {
  x = (1ll * x * y % mod + mod) % mod;
}
template <typename A>
inline void debug(A a) {
  cout << a << '\n';
}
template <typename A>
inline long long sqr(A x) {
  return 1ll * x * x;
}
template <typename A>
A inv(A x) {
  return mul(x, mod - 2);
}
template <typename A>
void bugl(A a[], int l, int r) {
  for (int i = l; i <= r; i++) cout << a[i] << " ";
  cout << endl;
}
template <typename A>
void bug(A a[], int l, int r) {
  for (int i = l; i <= r; i++) cout << a[i] << " ";
  cout << endl;
}
template <class T>
void wt(const T& x) {
  cout << x << endl;
}
template <class T>
void wt(const T& x, char c) {
  cout << x << c;
}
template <class T>
void wt(const T& x, const string& s) {
  cout << x << s;
}
template <class T>
void wt(const T& x, int rnd) {
  cout << fixed << setprecision(rnd) << x << endl;
}
inline long long lowbit(long long x) { return (x & (-x)); }
inline void debug() { cout << "###!!!" << endl; }
inline long long read() {
  char c = getchar();
  long long x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = 1ll * x * 10 + c - '0', c = getchar();
  return x * f;
}
long long n, k;
long long a[maxn];
int main() {
  string str1, str2;
  while (cin >> str1 >> str2) {
    std::vector<string> v;
    while (str1[0] < str2[0] && str1[1] < str2[1])
      v.push_back("RU"), str1[1]++, str1[0]++;
    while (str1[0] > str2[0] && str1[1] > str2[1])
      v.push_back("LD"), str1[0]--, str1[1]--;
    while (str1[0] > str2[0] && str1[1] < str2[1])
      v.push_back("LU"), str1[1]++, str1[0]--;
    while (str1[0] < str2[0] && str1[1] > str2[1])
      v.push_back("RD"), str1[0]++, str1[1]--;
    while (str1[0] != str2[0] && str1[0] < str2[0]) v.push_back("R"), str1[0]++;
    while (str1[0] != str2[0] && str1[0] > str2[0]) v.push_back("L"), str1[0]--;
    while (str1[1] != str2[1] && str1[1] < str2[1]) v.push_back("U"), str1[1]++;
    while (str1[1] != str2[1] && str1[1] > str2[1]) v.push_back("D"), str1[1]--;
    wt(v.size());
    for (auto d : v) {
      wt(d);
    }
  }
  return 0;
}