ABC 454 E

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

using ll = long long;

constexpr ll mod = 1e9+7;

void solve(){
    ll n, a, b; cin >> n >> a >> b;
    // if N odd or (A+B) even -> NOT POSSIBLE
    if ((n & 1) || !((a+b) & 1)) {cout << "No\n"; return;}
    cout << "Yes\n";

    // keep reducing boundaries by 4 possible ways ensuring top left is start and bottom right is end. Reduce until 2x2. Easy to solve 2x2.
    ll l = 1, r = n, u = 1, d = n;
    vector<string> st, ed;
    while (r - l > 1 || d - u > 1){
        if (a - u >= 2){                                                            // a
            st.push_back(string(r-l, 'R') + "D" + string(r-l, 'L') + "D");
            u += 2;
        } else if (d - a >= 2){                                                     // c
            ed.push_back("D" + string(r-l, 'L') + "D" + string(r-l, 'R'));
            d -= 2;    
        } else if (b - l >= 2){                                                      // b
            st.push_back(string(d-u, 'D') + "R" + string(d-u, 'U') + "R");
            l += 2;
        } else {                                                                    // d
            ed.push_back("R" + string(d-u, 'U') + "R" + string(d-u, 'D'));
            r -= 2;
        }
    }
    reverse(ed.begin(), ed.end());

    for (string& s : st) cout << s;
    cout << (a == u && b == r ? "DR" : "RD");
    for (string& s : ed) cout << s;
    cout << '\n';
}

int main(){
    ios_base::sync_with_stdio(false); cin.tie(nullptr);

    ll t; cin >> t;
    while (t--) solve();
}

// TC - O(N*N) due to moves
// SC - O(N*N) due to moves 
// N^2 - 2 moves