Problem: This is an interactive problem.

Bob lives in a square grid of size n × n, with rows numbered 1 through n from top to bottom, and columns numbered 1 through n from left to right. Every cell is either allowed or blocked, but you don't know the exact description of the grid. You are given only an integer n.

Bob can move through allowed cells but only in some limited directions. When Bob is in an allowed cell in the grid, he can move down or right to an adjacent cell, if it is allowed.

You can ask at most 4 ⋅ n queries of form "? r_1 c_1 r_2 c_2" (1 ≤ r_1 ≤ r_2 ≤ n, 1 ≤ c_1 ≤ c_2 ≤ n). The answer will be "YES" if Bob can get from a cell (r_1, c_1) to a cell (r_2, c_2), and "NO" otherwise. In particular, if one of the two cells (or both) is a blocked cell then the answer is "NO" for sure. Since Bob doesn't like short trips, you can only ask queries with the manhattan distance between the two cells at least n - 1, i.e. the following condition must be satisfied: (r_2 - r_1) + (c_2 - c_1) ≥ n - 1.

It's guaranteed that Bob can get from the top-left corner (1, 1) to the bottom-right corner (n, n) and your task is to find a way to do it. You should print the answer in form "! S" where S is a string of length 2 ⋅ n - 2 consisting of characters 'D' and 'R', denoting moves down and right respectively. The down move increases the first coordinate by 1, the right move increases the second coordinate by 1. If there are multiple solutions, any of them will be accepted. You should terminate immediately after printing the solution.

Input

The only line of the input contains an integer n (2 ≤ n ≤ 500) — the size of the grid.

Output

When you are ready to print the answer, print a single line containing "! S" where where S is a string of length 2 ⋅ n - 2 consisting of characters 'D' and 'R', denoting moves down and right respectively. The path should be a valid path going from the cell (1, 1) to the cell (n, n) passing only through allowed cells.

Interaction

You can ask at most 4 ⋅ n queries. To ask a query, print a line containing "? r_1 c_1 r_2 c_2" (1 ≤ r_1 ≤ r_2 ≤ n, 1 ≤ c_1 ≤ c_2 ≤ n). After that you should read a single line containing "YES" or "NO" depending on the answer of the query. "YES" means Bob can go from the cell (r_1, c_1) to the cell (r_2, c_2), "NO" means the opposite.

Note that the grid is fixed before the start of your solution and does not depend on your queries.

After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded or other negative verdict. To do this, use: 

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



Answer "BAD" instead of "YES" or "NO" means that you made an invalid query. Exit immediately after receiving "BAD" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.

Example

Input

4
 
YES
 
NO
 
YES
 
YES
 


Output

 
? 1 1 4 4
 
? 1 2 4 3
 
? 4 1 4 4
 
? 1 4 4 4
 
! RDRRDD

Note

The first example is shown on the picture below.

<image>

To hack, use the following input format:

The first line should contain a single integer n (2 ≤ n ≤ 500) — the size of the grid.

Each of the next n lines should contain a string of n characters '#' or '.', where '#' denotes a blocked cell, and '.' denotes an allowed cell.

For example, the following text encodes the example shown above:
    
    
      
    4  
    ..#.  
    #...  
    ###.  
    ....  
    

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1000000007;
long long int N, M, K, H, W, L, R;
class Segment_Tree {
  vector<long long int> v;
  vector<int> l;
  vector<int> r;
  int num;
  long long int ret;
  bool is_min;
  long long int Update(int place) {
    if (place >= v.size() / 2) {
      return v[place];
    }
    if (is_min) {
      v[place] = min(Update(place * 2), Update(place * 2 + 1));
      return v[place];
    }
    v[place] = max(Update(place * 2), Update(place * 2 + 1));
    return v[place];
  }
  void RMQ(int a, int b, int place) {
    if (l[place] >= a && r[place] <= b) {
      if (is_min)
        ret = min(ret, v[place]);
      else
        ret = max(ret, v[place]);
      return;
    }
    if (l[place] > b || r[place] < a) return;
    RMQ(a, b, place * 2);
    RMQ(a, b, place * 2 + 1);
    return;
  }

 public:
  void Left(int place) {
    if (place >= v.size() / 2) {
      l[place] = place - v.size() / 2;
      return;
    }
    Left(place * 2);
    Left(place * 2 + 1);
    l[place] = l[place * 2];
    return;
  }
  void Right(int place) {
    if (place >= v.size() / 2) {
      r[place] = place - v.size() / 2;
      return;
    }
    Right(place * 2);
    Right(place * 2 + 1);
    r[place] = r[place * 2 + 1];
    return;
  }
  Segment_Tree(int n, bool min) {
    n++;
    num = 1;
    while (num < n * 2) {
      num *= 2;
    }
    l.resize(num);
    r.resize(num);
    is_min = min;
    if (min) {
      v.resize(num, LLONG_MAX);
    } else
      v.resize(num, LLONG_MIN);
    Left(1);
    Right(1);
  }
  void Insert(int place, long long int num, bool update) {
    place += v.size() / 2;
    v[place] = num;
    if (!update) return;
    place /= 2;
    while (place) {
      if (is_min)
        v[place] = min(v[place * 2], v[place * 2 + 1]);
      else
        v[place] = max(v[place * 2], v[place * 2 + 1]);
      place /= 2;
    }
  }
  void TopDown() { Update(1); }
  long long int RMQ(int a, int b) {
    if (is_min)
      ret = LLONG_MAX;
    else
      ret = LLONG_MIN;
    RMQ(a, b, 1);
    return ret;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  int sx = 1, sy = 1, gx = N, gy = N;
  int cx = 1, cy = 1;
  string ans;
  string add;
  while (gx + gy - cx - cy >= N) {
    cout << "? " << cy + 1 << " " << cx << " " << gy << " " << gx << endl;
    string s;
    cin >> s;
    if (s == "YES") {
      ans.push_back('D');
      cy++;
    } else {
      ans.push_back('R');
      cx++;
    }
  }
  cy = N;
  cx = N;
  while (cy + cx - sx - sy >= N) {
    cout << "? " << sy << " " << sx << " " << cy << " " << cx - 1 << endl;
    string s;
    cin >> s;
    if (s == "YES") {
      add = 'R' + add;
      cx--;
    } else {
      add = 'D' + add;
      cy--;
    }
  }
  cout << "! " << ans << add << endl;
  return 0;
}