Problem: Bashar was practicing for the national programming contest. Because of sitting too much in front of the computer without doing physical movements and eating a lot Bashar became much fatter. Bashar is going to quit programming after the national contest and he is going to become an actor (just like his father), so he should lose weight.

In order to lose weight, Bashar is going to run for k kilometers. Bashar is going to run in a place that looks like a grid of n rows and m columns. In this grid there are two one-way roads of one-kilometer length between each pair of adjacent by side cells, one road is going from the first cell to the second one, and the other road is going from the second cell to the first one. So, there are exactly (4 n m - 2n - 2m) roads.

Let's take, for example, n = 3 and m = 4. In this case, there are 34 roads. It is the picture of this case (arrows describe roads):

<image>

Bashar wants to run by these rules:

  * He starts at the top-left cell in the grid; 
  * In one move Bashar may go up (the symbol 'U'), down (the symbol 'D'), left (the symbol 'L') or right (the symbol 'R'). More formally, if he stands in the cell in the row i and in the column j, i.e. in the cell (i, j) he will move to: 
    * in the case 'U' to the cell (i-1, j); 
    * in the case 'D' to the cell (i+1, j); 
    * in the case 'L' to the cell (i, j-1); 
    * in the case 'R' to the cell (i, j+1); 
  * He wants to run exactly k kilometers, so he wants to make exactly k moves; 
  * Bashar can finish in any cell of the grid; 
  * He can't go out of the grid so at any moment of the time he should be on some cell; 
  * Bashar doesn't want to get bored while running so he must not visit the same road twice. But he can visit the same cell any number of times. 



Bashar asks you if it is possible to run by such rules. If it is possible, you should tell him how should he run.

You should give him a steps to do and since Bashar can't remember too many steps, a should not exceed 3000. In every step, you should give him an integer f and a string of moves s of length at most 4 which means that he should repeat the moves in the string s for f times. He will perform the steps in the order you print them.

For example, if the steps are 2 RUD, 3 UUL then the moves he is going to move are RUD + RUD + UUL + UUL + UUL = RUDRUDUULUULUUL.

Can you help him and give him a correct sequence of moves such that the total distance he will run is equal to k kilometers or say, that it is impossible?

Input

The only line contains three integers n, m and k (1 ≤ n, m ≤ 500, 1 ≤ k ≤ 10 ^{9}), which are the number of rows and the number of columns in the grid and the total distance Bashar wants to run.

Output

If there is no possible way to run k kilometers, print "NO" (without quotes), otherwise print "YES" (without quotes) in the first line.

If the answer is "YES", on the second line print an integer a (1 ≤ a ≤ 3000) — the number of steps, then print a lines describing the steps.

To describe a step, print an integer f (1 ≤ f ≤ 10^{9}) and a string of moves s of length at most 4. Every character in s should be 'U', 'D', 'L' or 'R'.

Bashar will start from the top-left cell. Make sure to move exactly k moves without visiting the same road twice and without going outside the grid. He can finish at any cell.

We can show that if it is possible to run exactly k kilometers, then it is possible to describe the path under such output constraints.

Examples

Input


3 3 4


Output


YES
2
2 R
2 L


Input


3 3 1000000000


Output


NO


Input


3 3 8


Output


YES
3
2 R
2 D
1 LLRR


Input


4 4 9


Output


YES
1
3 RLD


Input


3 4 16


Output


YES
8
3 R
3 L
1 D
3 R
1 D
1 U
3 L
1 D

Note

The moves Bashar is going to move in the first example are: "RRLL".

It is not possible to run 1000000000 kilometers in the second example because the total length of the roads is smaller and Bashar can't run the same road twice.

The moves Bashar is going to move in the third example are: "RRDDLLRR".

The moves Bashar is going to move in the fifth example are: "RRRLLLDRRRDULLLD". It is the picture of his run (the roads on this way are marked with red and numbered in the order of his running):

<image>

Solution (in JAVA):

Solution:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.Reader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class D {
    public static void main(String[] args) {
        var scanner = new BufferedScanner();
        var writer = new PrintWriter(new BufferedOutputStream(System.out));

        var n = scanner.nextInt();
        var m = scanner.nextInt();
        var k = scanner.nextInt();
        var ans = solve(n, m, k);
        if (ans != null) {
            writer.println("YES");
            writer.println(ans.size());
            assert ans.size() <= 3000;
            for (Step step : ans) {
                writer.println(step.f + " " + step.s);
            }
        } else {
            writer.println("NO");
        }

        scanner.close();
        writer.flush();
        writer.close();
    }

    static final int[][] MOVE = {
            {0, 1}, // R
            {1, 0}, // D
            {-1, 0}, // U
            {0, -1} // L
    };
    static final char[] DIR = {0, 'R', 'D', 'U', 'L'};

    private static List<Step> solve(int n, int m, int k) {
        var atMost = 4 * n * m - 2 * n - 2 * m;
        if (k > atMost) {
            return null;
        }
        // 每条路只能走一遍，每个格子可以走多遍
        // atMost步是一定可以走通的。主要问题变成了怎么分组使得能够在3000步以内表达。
        var ans = new ArrayList<Step>();
        var current = 0;
        var r = 1;
        var c = 1;
        while (current < k) {
            // RDU...
            if (r < n) {
                var countRDU = Math.min(m - c, (k - current) / 3);
                if (countRDU > 0) {
                    ans.add(new Step(countRDU, "RDU"));
                    c += countRDU;
                    current += 3 * countRDU;
                    if (current == k) {
                        break;
                    }
                    if (k - current < 3 && c < m) {
//                    finishRDU(k - current, ans);
                        if (k - current == 2) {
                            ans.add(new Step(1, "R"));
                            ans.add(new Step(1, "D"));
                        } else {
                            ans.add(new Step(1, "R"));
                        }
                        break;
                    }
                }
            } else { // R...
                var countR = Math.min(k - current, m - c);
                if (countR > 0) {
                    ans.add(new Step(countR, "R"));
                    c += countR;
                    current += countR;
                    if (current == k) {
                        break;
                    }
                }
            }
            // L...
            {
                var countL = Math.min(k - current, c - 1);
                if (countL > 0) {
                    ans.add(new Step(countL, "L"));
                    c -= countL;
                    current += countL;
                    if (current == k) {
                        break;
                    }
                }
            }
            // D...
            if (r < n) {
                ans.add(new Step(1, "D"));
                r++;
                current++;
                if (current == k) {
                    break;
                }
            } else {
                var countU = k - current;
                ans.add(new Step(countU, "U"));
                r -= countU;
                current += countU;
            }
        }
        return ans;
    }

    static final boolean DEBUG = true;

    private static void log(Object... a) {
        if (DEBUG) {
            for (Object x : a) {
                System.out.println(x + " ");
            }
        }
    }

    static class Step {
        int f;
        String s;

        Step(int f, String s) {
            this.f = f;
            this.s = s;
            assert s.length() <= 4;
        }
    }

    public static class BufferedScanner {
        BufferedReader br;
        StringTokenizer st;

        public BufferedScanner(Reader reader) {
            br = new BufferedReader(reader);
        }

        public BufferedScanner() {
            this(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        void close() {
            try {
                br.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

    }

    static long gcd(long a, long b) {
        if (a < b) {
            return gcd(b, a);
        }
        while (b > 0) {
            long tmp = b;
            b = a % b;
            a = tmp;
        }
        return a;
    }

    static long inverse(long a, long m) {
        long[] ans = extgcd(a, m);
        return ans[0] == 1 ? (ans[1] + m) % m : -1;
    }

    private static long[] extgcd(long a, long m) {
        if (m == 0) {
            return new long[]{a, 1, 0};
        } else {
            long[] ans = extgcd(m, a % m);
            long tmp = ans[1];
            ans[1] = ans[2];
            ans[2] = tmp;
            ans[2] -= ans[1] * (a / m);
            return ans;
        }
    }

    private static List<Integer> primes(double upperBound) {
        var limit = (int) Math.sqrt(upperBound);
        var isComposite = new boolean[limit + 1];
        var primes = new ArrayList<Integer>();
        for (int i = 2; i <= limit; i++) {
            if (isComposite[i]) {
                continue;
            }
            primes.add(i);
            int j = i + i;
            while (j <= limit) {
                isComposite[j] = true;
                j += i;
            }
        }
        return primes;
    }


}