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.*;
import java.util.*;
import java.lang.*;

public class Main {
    final static String fileName = "";
    final static boolean useFiles = false;

    public static void main(String[] args) throws FileNotFoundException {
        InputStream inputStream;
        OutputStream outputStream;
        if (useFiles) {
            inputStream = new FileInputStream(fileName + ".in");
            outputStream = new FileOutputStream(fileName + ".out");
        } else {
            inputStream = System.in;
            outputStream = System.out;
        }
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task(in, out);
        solver.solve();
        out.close();
    }
}

class STree {
    long[] sum, add;
    int n;

    STree(int n) {
        sum = new long[Integer.highestOneBit(n) << 2];
        add = new long[Integer.highestOneBit(n) << 2];
        this.n = n;
    }

    long get(int v, int tl, int tr, int l, int r, long delta) {
        if (l > r)
            return 0;
        if (tl == l && tr == r)
            return sum[v] + delta * (tr - tl + 1);
        int tm = (tl + tr) / 2;
        delta += add[v];
        return get(2 * v + 1, tl, tm, l, Math.min(r, tm), delta)
                + get(2 * v + 2, tm + 1, tr, Math.max(l, tm + 1), r, delta);
    }

    void set(int v, int tl, int tr, int l, int r, long value) {
        if (l > r)
            return;
        if (tl == l && tr == r) {
            add[v] += value;
            sum[v] += (tr - tl + 1) * value;
        }
        else {
            int tm = (tl + tr) / 2;
            set(2 * v + 1, tl, tm, l, Math.min(r, tm), value);
            set(2 * v + 2, tm + 1, tr, Math.max(l, tm + 1), r, value);
            sum[v] = sum[2 * v + 1] + sum[2 * v + 2] + (tr - tl + 1) * add[v];
        }
    }
}

class Task {
    public void solve() {
        int n = in.nextInt(), m = in.nextInt(), w = in.nextInt();
        STree x = new STree(n), y = new STree(m);
        long all = 0;
        for (int i = 0; i < w; i++) {
            int type = in.nextInt();
            int x1 = in.nextInt() - 1, y1 = in.nextInt() - 1, x2 = in.nextInt() - 1, y2 = in.nextInt() - 1;
            if (type == 0) {
                long v = in.nextLong();
                x.set(0, 0, n - 1, x1, x2, v * (y2 - y1 + 1));
                y.set(0, 0, m - 1, y1, y2, v * (x2 - x1 + 1));
                all += (y2 - y1 + 1) * v * (x2 - x1 + 1);
            } else {
                long result = x.get(0, 0, n - 1, x1, x2, 0)
                        + y.get(0, 0, m - 1, y1, y2, 0)
                        - all;
                out.println(result);
            }
        }
    }

    private InputReader in;
    private PrintWriter out;

    Task(InputReader in, PrintWriter out) {
        this.in = in;
        this.out = out;
    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream), 32768);
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public double nextDouble(){
        return Double.parseDouble(next());
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public long nextLong(){
        return Long.parseLong(next());
    }
}