Problem: This is an interactive problem.

Vladik has favorite game, in which he plays all his free time.

Game field could be represented as n × m matrix which consists of cells of three types: 

  * «.» — normal cell, player can visit it. 
  * «F» — finish cell, player has to finish his way there to win. There is exactly one cell of this type. 
  * «*» — dangerous cell, if player comes to this cell, he loses. 



Initially player is located in the left top cell with coordinates (1, 1). 

Player has access to 4 buttons "U", "D", "L", "R", each of them move player up, down, left and right directions respectively.

But it’s not that easy! Sometimes friends play game and change functions of buttons. Function of buttons "L" and "R" could have been swapped, also functions of buttons "U" and "D" could have been swapped. Note that functions of buttons can be changed only at the beginning of the game.

Help Vladik win the game!

Input

First line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — number of rows and columns respectively.

Each of next n lines contains m characters describing corresponding row of field. Set of characters in field is described above.

Guaranteed that cell with coordinates (1, 1) is normal and there is at least one way from initial cell to finish cell without dangerous cells. 

Interaction

You can press buttons no more than 2·n·m times.

To press a button you should print "U", "D", "L", "R" in new line. It’s necessary to print newline character and flush output. After flushing buffer you should read answer from input data. Answer is the pair of space-separated integers x, y — new position of player. In case, if there is no cell in direction of moving, position will not change. If after any move player lost, in other words player move to dangerous cell, then x and y will be equal to  - 1.

If after any move player is in finish or dangerous cell, then you should terminate your program.

To finish output buffer (i. e. for operation flush) right after printing direction and newline you should do next:

  * fflush(stdout) in C++ 
  * System.out.flush() in Java 
  * stdout.flush() in Python 
  * flush(output) in Pascal 
  * read documentation for other languages. 



Hacks

To perform a hack you should use this format:
    
    
      
    n m swapLR swapUD    
    a_1    
    a_2    
    ...    
    a_n  
    

Where n, m — number of rows and columns in game field. swapLR is equal to 1 in case, when directions "L’’ and "R’’ is swapped, and equal to 0 otherwise. swapUD is equal to 1, when directions "U’’ and "D’’ is swapped, and equal to 0 otherwise. a1, a2, ..., an — description of corresponding rows of game field.

Example

Input

4 3
...
**.
F*.
...
1 1
1 2
1 3
1 3
2 3
3 3
4 3
4 2
4 1
3 1


Output

R
L
L
D
U
U
U
R
R
D

Note

In first test case all four directions swapped with their opposite directions. Protocol of interaction In more convenient form:

<image>

This test could be presenter for hack in following way: 
    
    
      
    4 3 1 1  
    ...  
    **.  
    F*.  
    ...  
    

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int inf = 1000000000;
const long long int mod = 1000000000 + 7;
inline void IO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
inline int dcmp(long double x) { return x < -1e-12 ? -1 : (x > 1e-12); }
template <class T>
inline int CHECK(T MASK, int i) {
  return (MASK >> i) & 1;
}
template <class T>
inline T ON(T MASK, int i) {
  return MASK | (T(1) << i);
}
template <class T>
inline T OFF(T MASK, int i) {
  return MASK & (~(T(1) << i));
}
template <typename T>
inline int CNT(T MASK) {
  if (numeric_limits<T>::digits <= numeric_limits<unsigned int>::digits)
    return __builtin_popcount(MASK);
  else
    return __builtin_popcountll(MASK);
}
template <class T>
inline int RIGHT(T MASK) {
  return log2(MASK & -MASK);
}
int dx4[] = {0, 0, -1, +1};
int dy4[] = {+1, -1, 0, 0};
int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1, 0};
int dy8[] = {0, 1, 1, 1, 0, -1, -1, -1, 0};
inline void I(int& a) { scanf("%d", &a); }
inline void I(long long int& a) { scanf("%I64d", &a); }
inline void I(unsigned long long int& a) { scanf("%I64u", &a); }
inline void I(char* a) { scanf("%s", a); }
char Iarr[2000010];
inline void I(string& a) {
  scanf("%s", Iarr);
  a = Iarr;
}
template <typename T, typename... Args>
void I(T& a, Args&... args) {
  I(a);
  I(args...);
}
inline void OUT(int a) { printf("%d", a); }
inline void OUT(long long int a) { printf("%I64d", a); }
inline void OUT(const char* a) { printf("%s", a); }
inline void OUT(char* a) { printf("%s", a); }
inline void OUT(bool a) { printf("%d", a); }
inline void OUT(string a) {
  for (__typeof(a.end()) it = (a.begin()) - ((a.begin()) > (a.end()));
       it != (a.end()) - ((a.begin()) > (a.end()));
       it += 1 - 2 * ((a.begin()) > (a.end())))
    printf("%c", *it);
}
inline void OUT(unsigned long long int a) { printf("%I64u", a); }
template <typename T, typename... Args>
void OUT(T a, Args... args) {
  OUT(a);
  OUT(" ");
  OUT(args...);
}
template <typename... Args>
void O(Args... args) {
  OUT(args...);
  OUT("\n");
}
template <typename T1, typename T2>
inline ostream& operator<<(ostream& os, pair<T1, T2> p) {
  os << "{" << p.first << ", " << p.second << "}";
  return os;
}
template <typename T>
inline ostream& operator<<(ostream& os, vector<T>& a) {
  os << "[";
  for (int i = 0; i < (int)a.size(); i++) {
    if (i) os << ", ";
    os << a[i];
  }
  os << "]";
  return os;
}
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cout << *it << " = " << a << endl;
  err(++it, args...);
}
const int M = 110;
int n, m;
bool vis[M][M];
pair<int, int> par[M][M];
string s[M];
vector<pair<int, int> > p;
void dfs(int x, int y, int px, int py) {
  vis[x][y] = 1;
  par[x][y] = make_pair(px, py);
  for (int i = 0; i < 4; i++) {
    int nx = x + dx4[i];
    int ny = y + dy4[i];
    if (nx >= 0 && nx < n && ny >= 0 && ny < m && !vis[nx][ny]) {
      if (s[nx][ny] != '*') dfs(nx, ny, x, y);
    }
  }
}
void path(int x, int y) {
  if (x || y) {
    path(par[x][y].first, par[x][y].second);
  }
  p.push_back(make_pair(x, y));
}
int main() {
  I(n, m);
  for (int i = 0; i < n; i++) {
    I(s[i]);
  }
  dfs(0, 0, -1, -1);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (s[i][j] == 'F') {
        path(i, j);
      }
    }
  }
  int i = 0;
  int N = (int)p.size();
  bitset<2> B;
  B.reset();
  bool D[2] = {0};
  while (i != N - 1) {
    if (p[i + 1].second > p[i].second || p[i + 1].second < p[i].second) {
      if (!D[0]) {
        if (p[i + 1].second > p[i].second) {
          cout << "R" << endl;
        } else {
          cout << "L" << endl;
        }
        int x, y;
        cin >> x >> y;
        x--;
        y--;
        if (make_pair(x, y) == p[i]) {
          B[0] = 1;
          if (p[i + 1].second > p[i].second) {
            cout << "L" << endl;
          } else {
            cout << "R" << endl;
          }
          cin >> x >> y;
        }
        D[0] = 1;
      } else {
        if (!B[0]) {
          if (p[i + 1].second > p[i].second) {
            cout << "R" << endl;
          } else {
            cout << "L" << endl;
          }
        } else {
          if (p[i + 1].second < p[i].second) {
            cout << "R" << endl;
          } else {
            cout << "L" << endl;
          }
        }
        int x, y;
        cin >> x >> y;
      }
    } else {
      if (!D[1]) {
        if (p[i + 1].first > p[i].first) {
          cout << "D" << endl;
        } else {
          cout << "U" << endl;
        }
        int x, y;
        cin >> x >> y;
        x--;
        y--;
        if (make_pair(x, y) == p[i]) {
          B[1] = 1;
          if (p[i + 1].first > p[i].first) {
            cout << "U" << endl;
          } else {
            cout << "D" << endl;
          }
          cin >> x >> y;
        }
        D[1] = 1;
      } else {
        if (!B[1]) {
          if (p[i + 1].first > p[i].first) {
            cout << "D" << endl;
          } else {
            cout << "U" << endl;
          }
        } else {
          if (p[i + 1].first < p[i].first) {
            cout << "D" << endl;
          } else {
            cout << "U" << endl;
          }
        }
        int x, y;
        cin >> x >> y;
      }
    }
    i++;
  }
  return 0;
}