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.io.*;

public class Main {
	static istream fi = new istream ();
	static ostream fo = new ostream (0);
	static long[] C, R, UC, UR;
	static int m, n;
	
	static void update (long[] T,long[] U,int x,int l,int r,int u,int v,long p) {
		if (r < u || l > v) return;
		if (u <= l && r <= v) {
			T[x] += p * (r - l + 1);
			U[x] += p;
			return;
		}
		int mid = (l + r) >> 1, k = x << 1;
		if (U[x] > 0) {
			U[k] += U[x];
			T[k] += U[x] * (mid - l + 1);
			U[k + 1] += U[x];
			T[k + 1] += U[x] * (r - mid);
			U[x] = 0;
		}
		update(T, U, k, l, mid, u, v, p);
		update(T, U, k + 1, mid + 1, r, u, v, p);
		T[x] = T[k] + T[k + 1];
	}
	
	static long get (long[] T,long[] U,int x,int l,int r,int u,int v) {
		if (r < u || l > v) return 0L;
		if (u <= l && r <= v) return T[x];
		int mid = (l + r) >> 1, k = x << 1;
		if (U[x] > 0) {
			U[k] += U[x];
			T[k] += U[x] * (mid - l + 1);
			U[k + 1] += U[x];
			T[k + 1] += U[x] * (r - mid);
			U[x] = 0;
		}
		return get(T, U, k, l, mid, u, v) + get(T, U, k + 1, mid + 1, r, u, v);
	}

	public static void main(String[] args) {
		m = fi.nextInt();
		n = fi.nextInt();
		C = new long [Integer.highestOneBit(n) << 2];
		UC = new long [Integer.highestOneBit(n) << 2];
		R = new long [Integer.highestOneBit(m) << 2];
		UR = new long [Integer.highestOneBit(m) << 2];
		int q = fi.nextInt(), t, x, y, u, v;
		long p;
		
		while (q-- > 0) {
			t = fi.nextInt();
			x = fi.nextInt();
			y = fi.nextInt();
			u = fi.nextInt();
			v = fi.nextInt();
			if (t == 0) {
				p = fi.nextInt();
				update(R, UR, 1, 1, m, x, u, p * (v - y + 1));
				update(C, UC, 1, 1, n, y, v, p * (u - x + 1));
			} else {
				p = get(R, UR, 1, 1, m, x, u);
				p+= get(C, UC, 1, 1, n, y, v);
				fo.print((p - R[1]) + "\n");
			}
		}
	    fo.close();
	}
}

class istream {
	BufferedInputStream fi = new BufferedInputStream (System.in);
	
	int read () {
		int k = 0;
		try {k = fi.read();} catch (IOException e) {}
		return k;
	}
	
	int nextInt () {
		int c = 0, k;
		while ((k = read()) < '0') {}
		while (k >= '0') {
			c = c * 10 + k - '0';
			k = read();
		}
		return c;
	}
}

class ostream extends PrintWriter {
	public ostream (int x) {
		super (new BufferedOutputStream (System.out));
	}
}