Problem:
Score : 800 points

Problem Statement

There is a grid with R rows and C columns.
We call the cell in the r-th row and c-th column (r，c).

Mr. Takahashi wrote non-negative integers into N of the cells, that is, he wrote a non-negative integer a_i into (r_i，c_i) for each i (1≤i≤N).
After that he fell asleep.

Mr. Aoki found the grid and tries to surprise Mr. Takahashi by writing integers into all remaining cells.
The grid must meet the following conditions to really surprise Mr. Takahashi.

Condition 1: Each cell contains a non-negative integer.

Condition 2: For any 2×2 square formed by cells on the grid, the sum of the top left and bottom right integers must always equal to the sum of the top right and bottom left integers.

Determine whether it is possible to meet those conditions by properly writing integers into all remaining cells.

Constraints

2≤R，C≤10^5

1≤N≤10^5

1≤r_i≤R

1≤c_i≤C

(r_i，c_i) ≠ (r_j，c_j) (i≠j)

a_i is an integer.

0≤a_i≤10^9

Input

The input is given from Standard Input in the following format:

R C
N
r_1 c_1 a_1
r_2 c_2 a_2
:
r_N c_N a_N

Output

Print Yes if it is possible to meet the conditions by properly writing integers into all remaining cells.
Otherwise, print No.

Sample Input 1

2 2
3
1 1 0
1 2 10
2 1 20

Sample Output 1

Yes

You can write integers as follows.

Sample Input 2

2 3
5
1 1 0
1 2 10
1 3 20
2 1 30
2 3 40

Sample Output 2

No

There are two 2×2 squares on the grid, formed by the following cells:

Cells (1，1), (1，2), (2，1) and (2，2)

Cells (1，2), (1，3), (2，2) and (2，3)

You have to write 40 into the empty cell to meet the condition on the left square, but then it does not satisfy the condition on the right square.

Sample Input 3

2 2
3
1 1 20
1 2 10
2 1 0

Sample Output 3

No

You have to write -10 into the empty cell to meet condition 2, but then it does not satisfy condition 1.

Sample Input 4

3 3
4
1 1 0
1 3 10
3 1 10
3 3 20

Sample Output 4

Yes

You can write integers as follows.

Sample Input 5

2 2
4
1 1 0
1 2 10
2 1 30
2 2 20

Sample Output 5

No

All cells already contain a integer and condition 2 is not satisfied.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.io.InputStream;
import java.io.PrintStream;
import java.util.HashMap;
import java.util.NoSuchElementException;

public class Main {
	static FastScanner sc = new FastScanner();
	static PrintStream out = System.out;
	static HashMap<Long, Integer> map = new HashMap<Long, Integer>();

	public static void main(String args[]) {
		int R = ni();
		int C = ni();
		int N = ni();
		boolean check = true;
		int CD = -10000000;
		int RD = -10000000;
		int RR[] = new int[R];
		int Rmin[] = new int[R];
		int CC[] = new int[C];
		int Cmin[] = new int[C];
		for (int i=0;i<R;i++) {
			Rmin[i] = 1000000;
		}
		for (int i=0;i<C;i++) {
			Cmin[i] = 1000000;
		}

		for (int i=0;i<N;i++) {
			int r = ni() - 1;
			int c = ni() - 1;
			int a = ni();

			if (Rmin[r] != 1000000) {
				int s = (Rmin[r] - c);
				int t = RR[r] - a;
				if (t%s != 0) {
					check = false;
				} else {
					int dd = t/s;
					if (RD == -10000000) {
						RD = dd;
					} else if (RD != dd) {
						check = false;
					}
				}
			} else {
				Rmin[r] = c;
				RR[r] = a;
			}

			if (Cmin[c] != 1000000) {
				int s = (Cmin[c] - r);
				int t = CC[c] - a;
				if (t%s != 0) {
					check = false;
				} else {
					int dd = t/s;
					if (CD == -10000000) {
						CD = dd;
					} else if (CD != dd) {
						check = false;
					}
				}
			} else {
				Cmin[c] = r;
				CC[c] = a;
			}
		}

		// minus check
		for (int i=0;i<R;i++) {
			long pos = Rmin[i];
			long val = RR[i];
			long d = RD;
			if (pos == 1000000) continue;
			if (d == -10000000) continue;

			if (val - pos * d < 0) check = false;
			if (val + (R - 1 - pos) * d < 0) check = false;
		}
		for (int i=0;i<C;i++) {
			long pos = Cmin[i];
			long val = CC[i];
			long d = CD;
			if (pos == 1000000) continue;
			if (d == -10000000) continue;

			if (val - pos * d < 0) check = false;
			if (val + (C - 1 - pos) * d < 0) check = false;
		}

		System.out.println(check ? "Yes" : "No");
	}

	private static String next() {
		return sc.next();
	}

	private static int ni() {
		return sc.nextInt();
	}

	private static long nl() {
		return sc.nextLong();
	}

	private static int[] na(int n) {
		int[] a = new int[n];
		for(int i = 0;i < n;i++) a[i] = ni();
		return a;
	}

	private static long[] nal(int n) {
		long[] a = new long[n];
		for(int i = 0;i < n;i++) a[i] = ni();
		return a;
	}

	private static double nd() {
		return Double.parseDouble( next() );
	}
}

class FastScanner {
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;
	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		}else{
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}
	private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
	private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
	private void skipUnprintable() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;}
	public boolean hasNext() { skipUnprintable(); return hasNextByte();}
	public String next() {
		if (!hasNext()) throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while(isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	public int nextInt() {
		if (!hasNext()) throw new NoSuchElementException();
		int n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while(true){
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			}else if(b == -1 || !isPrintableChar(b)){
				return minus ? -n : n;
			}else{
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
	public long nextLong() {
		if (!hasNext()) throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while(true){
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			}else if(b == -1 || !isPrintableChar(b)){
				return minus ? -n : n;
			}else{
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
}