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.Reader;
import java.util.InputMismatchException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Writer;
import java.io.IOException;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Niyaz Nigmatullin
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		FastScanner in = new FastScanner(inputStream);
		FastPrinter out = new FastPrinter(outputStream);
		TaskE solver = new TaskE();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskE {
    public void solve(int testNumber, FastScanner in, FastPrinter out) {
        int n = in.nextInt();
        int m = in.nextInt();
        int w = in.nextInt();
        FenwickMultiSum fx = new FenwickMultiSum(n);
        FenwickMultiSum fy = new FenwickMultiSum(m);
        long all = 0;
        for (int i = 0; i < w; i++) {
            int type = in.nextInt();
            int x1 = in.nextInt() - 1;
            int y1 = in.nextInt() - 1;
            int x2 = in.nextInt();
            int y2 = in.nextInt();
            if (type == 0) {
                long v = in.nextInt();
                all += v * (x2 - x1) * (y2 - y1);
                fx.add(x1, x2, (y2 - y1) * v);
                fy.add(y1, y2, (x2 - x1) * v);
            } else {
                out.println(fx.getSum(x1, x2) + fy.getSum(y1, y2) - all);
            }
        }
    }
}

class FastScanner extends BufferedReader {

    public FastScanner(InputStream is) {
        super(new InputStreamReader(is));
    }

    public int read() {
        try {
            int ret = super.read();
//            if (isEOF && ret < 0) {
//                throw new InputMismatchException();
//            }
//            isEOF = ret == -1;
            return ret;
        } catch (IOException e) {
            throw new InputMismatchException();
        }
    }

    static boolean isWhiteSpace(int c) {
        return c >= 0 && c <= 32;
    }

    public int nextInt() {
        int c = read();
        while (isWhiteSpace(c)) {
            c = read();
        }
        int sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = read();
        }
        int ret = 0;
        while (c >= 0 && !isWhiteSpace(c)) {
            if (c < '0' || c > '9') {
                throw new NumberFormatException("digit expected " + (char) c
                        + " found");
            }
            ret = ret * 10 + c - '0';
            c = read();
        }
        return ret * sgn;
    }

    public String readLine() {
        try {
            return super.readLine();
        } catch (IOException e) {
            return null;
        }
    }

}

class FastPrinter extends PrintWriter {

    public FastPrinter(OutputStream out) {
        super(out);
    }


}

class FenwickMultiSum {
    long[] a;
    long[] b;

    public FenwickMultiSum(int n) {
        a = new long[n];
        b = new long[n];
    }

    public void add(int x, long d) {
        addSuffix(a, x, d);
        addSuffix(b, x, d * (1 - x));
    }

    private static void addSuffix(long[] a, int x, long d) {
        for (int i = x; i < a.length; i |= i + 1) {
            a[i] += d;
        }
    }

    public void add(int left, int right, long d) {
        add(left, d);
        add(right, -d);
    }

    public long getSum(int x) {
        return getPrefix(a, x) * x + getPrefix(b, x);
    }

    public long getSum(int left, int right) {
        return getSum(right - 1) - getSum(left - 1);
    }

    private static long getPrefix(long[] a, int x) {
        long ret = 0;
        for (int i = x; i >= 0; i = ((i + 1) & i) - 1) {
            ret += a[i];
        }
        return ret;
    }
}

