codeforces 631B B. Print Check

B. Print Check

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Kris works in a large company "Blake Technologies". As a best engineer of the company he was assigned a task to develop a printer that will be able to print horizontal and vertical strips. First prototype is already built and Kris wants to tests it. He wants you to implement the program that checks the result of the printing.

Printer works with a rectangular sheet of paper of size n × m. Consider the list as a table consisting of n rows and m columns. Rows are numbered from top to bottom with integers from 1 to n, while columns are numbered from left to right with integers from 1 to m. Initially, all cells are painted in color 0.

Your program has to support two operations:

1. Paint all cells in row ri in color ai;
2. Paint all cells in column ci in color ai.

If during some operation i there is a cell that have already been painted, the color of this cell also changes to ai.

Your program has to print the resulting table after k operation.

Input

The first line of the input contains three integers n, m and k (1  ≤  n,  m  ≤ 5000, n·m ≤ 100 000, 1 ≤ k ≤ 100 000) — the dimensions of the sheet and the number of operations, respectively.

Each of the next k lines contains the description of exactly one query:

• 1 ri ai (1 ≤ ri ≤ n, 1 ≤ ai ≤ 109), means that row ri is painted in color ai;
• 2 ci ai (1 ≤ ci ≤ m, 1 ≤ ai ≤ 109), means that column ci is painted in color ai.

Output

Print n lines containing m integers each — the resulting table after all operations are applied.

Examples

input

3 3 3
1 1 3
2 2 1
1 2 2

output

3 1 3 
2 2 2 
0 1 0 

input

5 3 5
1 1 1
1 3 1
1 5 1
2 1 1
2 3 1

output

1 1 1 
1 0 1 
1 1 1 
1 0 1 
1 1 1 

Note

The figure below shows all three operations for the first sample step by step. The cells that were painted on the corresponding step are marked gray.

题意：给你这些方格，然后让你涂颜色，问最后的状态是什么样的;

思路：后面涂的会覆盖前面涂的，所以从后往前确定,由于k>>n+m,所以很多都重复涂的，可以flag标记，涂过的就跳过,具体的见代码;

AC代码：

#include <bits/stdc++.h>
using namespace std;
int n,m,k;
const int N=5004;
int mp[N][N],flag1[N],flag2[N];
int a[100003],b[100003],c[100003];
int main()
{
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=k;i++)
    {
        scanf("%d%d%d",&a[i],&b[i],&c[i]);
    }
    for(int i=k;i>0;i--)
    {
        if(a[i]==1)
        {
            if(!flag1[b[i]])
            {
                for(int j=1;j<=m;j++)
                {
                   if(!mp[b[i]][j])mp[b[i]][j]=c[i];
                }
                flag1[b[i]]=1;
            }
        }
        else
        {
            if(!flag2[b[i]])
            {
                for(int j=1;j<=n;j++)
                {
                   if(!mp[j][b[i]]) mp[j][b[i]]=c[i];
                }
                flag2[b[i]]=1;
            }
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<m;j++)
        {
            printf("%d ",mp[i][j]);
        }
        printf("%d
",mp[i][m]);
    }
    return 0;
}