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;
inline int Read() {
  char ch = getchar();
  while (!((ch >= '0' && ch <= '9') || ch == '-')) ch = getchar();
  int first = 0, p = 1;
  if (ch == '-') p = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') first = first * 10 + ch - '0', ch = getchar();
  return first * p;
}
const int MAXN = 1000000 * 4 + 10;
int w;
long long first[MAXN], f[MAXN], second[MAXN], g[MAXN], Mashiro;
int lowbit(int first) { return first & -first; }
void Add(long long *a, long long *b, int first, long long v) {
  for (int i = first; i <= a[0]; i += lowbit(i))
    a[i] += v * (first - 1), b[i] += v;
}
long long Get(long long *a, long long *b, int first) {
  long long Tohka = 0;
  for (int i = first; i; i -= lowbit(i)) Tohka += b[i] * first - a[i];
  return Tohka;
}
int main() {
  first[0] = Read();
  second[0] = Read();
  w = Read();
  for (int i = 1; i <= w; ++i) {
    int ty, xl, yl, xr, yr, v;
    ty = Read();
    xl = Read();
    yl = Read();
    xr = Read();
    yr = Read();
    if (!ty)
      v = Read(), ++xr, ++yr, Mashiro += (long long)(xr - xl) * (yr - yl) * v,
      Add(first, f, xl, (long long)(yr - yl) * v),
      Add(first, f, xr, -(long long)(yr - yl) * v),
      Add(second, g, yl, (long long)(xr - xl) * v),
      Add(second, g, yr, -(long long)(xr - xl) * v);
    else
      printf("%I64d\n", Get(first, f, xr) - Get(first, f, xl - 1) +
                            Get(second, g, yr) - Get(second, g, yl - 1) -
                            Mashiro);
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}
