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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main229E {

    private static class FenwickTreeSumUpdateRange  {

        private final long[] array;
        private final long[] add;
        private final long[] sum;

        public FenwickTreeSumUpdateRange(int n) {
            array = new long[n];
            add = new long[n];
            sum = new long[n];
        }

        public void add(int left, int right, long value) {
            if(right < left) {
                throw new IllegalArgumentException();
            }
            add(right, value);
            add(left - 1, -value);
        }

        private void add(int right, long value) {
            if(right < 0) {
                return;
            }
            for(int i = right; i >= 0; i = (i & (i + 1)) - 1) {
                add[i] += value;
            }
            for(int i = right | (right + 1); i < array.length; i |= i + 1) {
                sum[i] += value * (right - (i & (i + 1)) + 1);
            }
        }

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

        private long sum(int right) {
            if(right < 0) {
                return 0;
            }
            long res = 0;
            for(int i = right; i >= 0; i = (i & (i + 1)) - 1) {
                res += sum[i] + add[i] * (i - (i & (i + 1)) + 1);
            }
            for(int i = right | (right + 1); i < array.length; i |= i + 1) {
                res += add[i] * (right - (i & (i + 1)) + 1);
            }
            return res;
        }

    }

    private FastScanner in;
    private PrintWriter out;

    public void solve() throws IOException {
        int n = in.nextInt();
        int m = in.nextInt();
        int w = in.nextInt();
        FenwickTreeSumUpdateRange treeR = new FenwickTreeSumUpdateRange(n);
        FenwickTreeSumUpdateRange treeC = new FenwickTreeSumUpdateRange(m);
        for(int i = 0; i < w; i++) {
            if(in.nextInt() == 0) {
                int x1 = in.nextInt() - 1;
                int y1 = in.nextInt() - 1;
                int x2 = in.nextInt() - 1;
                int y2 = in.nextInt() - 1;
                int v = in.nextInt();
                treeR.add(x1, x2, v * (y2 - y1 + 1));
                treeC.add(y1, y2, v * (x2 - x1 + 1));
            } else {
                int x1 = in.nextInt() - 1;
                int y1 = in.nextInt() - 1;
                int x2 = in.nextInt() - 1;
                int y2 = in.nextInt() - 1;
                out.println(treeR.sum(x1, x2) - treeC.sum(0, y1 - 1) -  treeC.sum(y2 + 1, m - 1));
            }
        }
    }

    public void run() {
        try {
            in = new FastScanner(System.in);
            out = new PrintWriter(System.out);
            solve();
            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    class FastScanner {

        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

    public static void main(String[] arg) {
        new Main229E().run();
    }
}