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;
struct ww {
  int ty, x, y, xx, yy, v;
} a[100010];
int i, j, k, n, nn, m, F;
long long an[100010], tr[2][4000010];
inline long long calc(int x, int y) {
  long long an = 0;
  for (; x; x -= x & -x) an += tr[y][x];
  return an;
}
inline long long cal(int x) { return calc(x - 1, 1) + calc(n - x + 1, 0) * x; }
inline void inc(int x, int y, long long z) {
  if (!x) return;
  for (; x <= n; x += x & -x) tr[y][x] += z;
}
inline void add(int x, long long y) { inc(n - x + 1, 0, y), inc(x, 1, y * x); }
void work() {
  memset(tr, 0, sizeof(tr));
  long long s = 0;
  for (i = 1; i <= m; i++)
    if (!a[i].ty) {
      long long A = (long long)(a[i].yy - a[i].y + 1) * a[i].v;
      add(a[i].xx, A), add(a[i].x - 1, -A);
      s += (long long)(a[i].xx - a[i].x + 1) * (a[i].yy - a[i].y + 1) * a[i].v;
    } else {
      if (F) an[i] -= s;
      an[i] += cal(a[i].xx) - cal(a[i].x - 1);
    }
}
int main() {
  scanf("%d%d%d", &n, &nn, &m);
  n = max(n, nn);
  for (i = 1; i <= m; i++) {
    scanf("%d%d%d%d%d", &a[i].ty, &a[i].x, &a[i].y, &a[i].xx, &a[i].yy);
    if (!a[i].ty) scanf("%d", &a[i].v);
  }
  work();
  F = 1;
  for (i = 1; i <= m; i++) swap(a[i].x, a[i].y), swap(a[i].xx, a[i].yy);
  work();
  for (i = 1; i <= m; i++)
    if (a[i].ty) printf("%I64d\n", an[i]);
  return 0;
}
