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.
#include <bits/stdc++.h>
using namespace std;
int gi() {
  int a;
  scanf("%d", &a);
  return a;
}
long long gli() {
  long long a;
  scanf("%I64d", &a);
  return a;
}
long long X[8388620];
long long XS[8388620];
long long Y[8388620];
long long YS[8388620];
void add(long long *A, long long *S, int i, int cx1, int cx2, int x1, int x2,
         long long v) {
  if (x1 <= cx1 && x2 >= cx2)
    A[i] += v;
  else if (x2 >= cx1 && x1 <= cx2) {
    S[i] += v * ((cx2 < x2 ? cx2 : x2) - (cx1 > x1 ? cx1 : x1) + 1);
    int m = (cx1 + cx2) / 2;
    add(A, S, 2 * i + 1, cx1, m, x1, x2, v);
    add(A, S, 2 * i + 2, m + 1, cx2, x1, x2, v);
  }
}
long long getv(long long *A, long long *S, int i, int cx1, int cx2, int x1,
               int x2) {
  if (x1 <= cx1 && x2 >= cx2) return S[i] + A[i] * (cx2 - cx1 + 1);
  if (x2 >= cx1 && x1 <= cx2) {
    int m = (cx1 + cx2) / 2;
    long long res = A[i] * ((cx2 < x2 ? cx2 : x2) - (cx1 > x1 ? cx1 : x1) + 1);
    long long vl = getv(A, S, 2 * i + 1, cx1, m, x1, x2);
    long long vr = getv(A, S, 2 * i + 2, m + 1, cx2, x1, x2);
    return res + vl + vr;
  }
  return 0;
}
int main() {
  int n = gi();
  int m = gi();
  memset(X, 0, sizeof(X));
  memset(XS, 0, sizeof(X));
  memset(Y, 0, sizeof(Y));
  memset(YS, 0, sizeof(Y));
  int q = gi();
  long long T = 0;
  for (int i = 0; i < q; i++) {
    int t = gi();
    int x1 = gi();
    int y1 = gi();
    int x2 = gi();
    int y2 = gi();
    if (t == 0) {
      long long v = gi();
      add(X, XS, 0, 1, n, x1, x2, v * (y2 - y1 + 1));
      add(Y, YS, 0, 1, m, y1, y2, v * (x2 - x1 + 1));
      T += v * (x2 - x1 + 1) * (y2 - y1 + 1);
    } else {
      long long vx = getv(X, XS, 0, 1, n, x1, x2);
      long long vy = getv(Y, YS, 0, 1, m, y1, y2);
      long long res = vx + vy - T;
      printf("%I64d\n", res);
    }
  }
  return 0;
}
