Inna loves sweets very much. That's why she wants to play the "Sweet Matrix" game with Dima and Sereja. But Sereja is a large person, so the game proved small for him. Sereja suggested playing the "Large Sweet Matrix" game.

The "Large Sweet Matrix" playing field is an n × m matrix. Let's number the rows of the matrix from 1 to n, and the columns — from 1 to m. Let's denote the cell in the i-th row and j-th column as (i, j). Each cell of the matrix can contain multiple candies, initially all cells are empty. The game goes in w moves, during each move one of the two following events occurs:

  1. Sereja chooses five integers x1, y1, x2, y2, v (x1 ≤ x2, y1 ≤ y2) and adds v candies to each matrix cell (i, j) (x1 ≤ i ≤ x2; y1 ≤ j ≤ y2). 
  2. Sereja chooses four integers x1, y1, x2, y2 (x1 ≤ x2, y1 ≤ y2). Then he asks Dima to calculate the total number of candies in cells (i, j) (x1 ≤ i ≤ x2; y1 ≤ j ≤ y2) and he asks Inna to calculate the total number of candies in the cells of matrix (p, q), which meet the following logical criteria: (p < x1 OR p > x2) AND (q < y1 OR q > y2). Finally, Sereja asks to write down the difference between the number Dima has calculated and the number Inna has calculated (D - I). 



Unfortunately, Sereja's matrix is really huge. That's why Inna and Dima aren't coping with the calculating. Help them!

Input

The first line of the input contains three integers n, m and w (3 ≤ n, m ≤ 4·106; 1 ≤ w ≤ 105).

The next w lines describe the moves that were made in the game. 

  * A line that describes an event of the first type contains 6 integers: 0, x1, y1, x2, y2 and v (1 ≤ x1 ≤ x2 ≤ n; 1 ≤ y1 ≤ y2 ≤ m; 1 ≤ v ≤ 109). 
  * A line that describes an event of the second type contains 5 integers: 1, x1, y1, x2, y2 (2 ≤ x1 ≤ x2 ≤ n - 1; 2 ≤ y1 ≤ y2 ≤ m - 1). 



It is guaranteed that the second type move occurs at least once. It is guaranteed that a single operation will not add more than 109 candies.

Be careful, the constraints are very large, so please use optimal data structures. Max-tests will be in pretests.

Output

For each second type move print a single integer on a single line — the difference between Dima and Inna's numbers.

Examples

Input

4 5 5
0 1 1 2 3 2
0 2 2 3 3 3
0 1 5 4 5 1
1 2 3 3 4
1 3 4 3 4


Output

2
-21

Note

Note to the sample. After the first query the matrix looks as: 
    
    
      
    22200  
    22200   
    00000  
    00000  
    

After the second one it is: 
    
    
      
    22200  
    25500  
    03300   
    00000  
    

After the third one it is: 
    
    
      
    22201  
    25501  
    03301  
    00001  
    

For the fourth query, Dima's sum equals 5 + 0 + 3 + 0 = 8 and Inna's sum equals 4 + 1 + 0 + 1 = 6. The answer to the query equals 8 - 6 = 2. For the fifth query, Dima's sum equals 0 and Inna's sum equals 18 + 2 + 0 + 1 = 21. The answer to the query is 0 - 21 = -21.
import java.util.*;
import java.io.*;

public class Solver{

    BufferedReader in;
    StringTokenizer str = null;
    PrintWriter out;

    private String next() throws Exception{
    	while (str == null || !str.hasMoreElements())
    	    str = new StringTokenizer(in.readLine());
    	return str.nextToken();
    }
    
    private int nextInt() throws Exception{
	return Integer.parseInt(next());
    }
    
    private long nextLong() throws Exception{
	return Long.parseLong(next());
    }
    
    public void run() throws Exception{
    	in = new BufferedReader(new InputStreamReader(System.in));
    	out = new PrintWriter(System.out);
	
	int n = nextInt(), m = nextInt(), w = nextInt();
	LazySegmentTree row = new LazySegmentTree(n);
	LazySegmentTree col = new LazySegmentTree(m);
	
	long total = 0;
	for(int i=0;i<w;i++){
	    int op = nextInt();
	    int x1 = nextInt()-1, y1 = nextInt()-1, x2 = nextInt()-1, y2 = nextInt()-1;
	    if (op == 0){
		long v = nextLong();
		row.update(x1, x2, (y2 - y1 + 1) * v);
		col.update(y1, y2, (x2 - x1 + 1) * v);
		total+=(y2 - y1 + 1)*(x2 - x1 + 1) * v;
	    }else{
		long s = row.sum(x1, x2) + col.sum(y1, y2);
		s-=total;
		
		out.println(s);
	    }
	}
	
        out.close();
    }
    
    static class LazySegmentTree{
	long []sum, add;
	int n;
	public LazySegmentTree(int len){
	    n = 1;
	    while(n < len) n<<=1;
	    sum = new long[2*n];
	    add = new long[2*n];
	    this.n = n;
	}
	
	public void update(int l, int r, long d){
	    update(1, 0, n-1, l, r, d);
	}
	
	public long sum(int l, int r){
	    return sum(1, 0, n-1, l, r);
	}
	
	private void push(int v, int l, int r){
	    if (add[v] != 0){
		int m = (l + r)/2;
		add[2*v]+=add[v];
		sum[2*v]+=(m - l + 1) * add[v];
		add[2*v+1]+=add[v];
		sum[2*v+1]+=(r - m) * add[v];
	    }
	    add[v] = 0;
	}
	
	private void update(int v, int tl, int tr, int l, int r, long d){
	    if (l > r) return;
	    if (tl == l && tr == r){
		add[v]+=d;
		sum[v]+=(r - l + 1) * d;
		return;
	    }
	    
	    push(v, tl, tr);
	    
	    int tm = (tl + tr)/2;
	    update(2*v, tl, tm, l, Math.min(tm, r), d);
	    update(2*v+1, tm+1, tr, Math.max(l, tm+1), r, d);
	    
	    sum[v] = sum[2*v] + sum[2*v+1];
	}
	
	private long sum(int v, int tl, int tr, int l, int r){
	    if (l > r) return 0;
	    if (tl == l && tr == r){
		return sum[v];
	    }
	    
	    push(v, tl, tr);
	    
	    int tm = (tl + tr)/2;
	    long left = sum(2*v, tl, tm, l, Math.min(tm, r));
	    long right = sum(2*v+1, tm+1, tr, Math.max(l, tm+1), r);
	    
	    return left + right;
	}
    }
    
    public static void main(String args[]) throws Exception{
	new Solver().run();
    }
}