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;
const int tope = 1 << 22;
void insertar(int pos, long long int val, long long int input[3][tope]) {
  if (pos == 0) return;
  int posini = pos;
  long long int(&valor)[tope] = input[0];
  long long int(&suma)[tope] = input[1];
  long long int(&sumaprop)[tope] = input[2];
  for (; pos > 0; pos -= pos & -pos) valor[pos] += val;
  pos = posini;
  sumaprop[pos] += val * posini;
  for (; pos < tope; pos += pos & -pos) suma[pos] += val * posini;
}
void insertarentre(int posinf, int possup, long long int val,
                   long long int input[3][tope]) {
  insertar(possup, val, input);
  insertar(posinf - 1, -val, input);
}
long long int computa(int pos, long long int input[3][tope]) {
  if (pos == 0) return 0;
  int posini = pos;
  long long int val = 0;
  long long int(&valor)[tope] = input[0];
  long long int(&suma)[tope] = input[1];
  long long int(&sumaprop)[tope] = input[2];
  val -= sumaprop[pos];
  for (; pos > 0; pos -= pos & -pos) val += suma[pos];
  pos = posini;
  for (; pos < tope; pos += pos & -pos) val += valor[pos] * posini;
  return val;
}
long long int computa(int posinf, int possup, long long int input[3][tope]) {
  return computa(possup, input) - computa(posinf - 1, input);
}
int n, m, w;
long long int fila[3][tope], columna[3][tope];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> w;
  for (int i = 0; i < w; i++) {
    int tipo;
    cin >> tipo;
    if (tipo == 0) {
      int i1, j1, i2, j2;
      long long int v;
      cin >> i1 >> j1 >> i2 >> j2 >> v;
      insertarentre(i1, i2, v * (j2 - j1 + 1), fila);
      insertarentre(j1, j2, v * (i2 - i1 + 1), columna);
    } else {
      int i1, j1, i2, j2;
      cin >> i1 >> j1 >> i2 >> j2;
      cout << -(computa(1, i1 - 1, fila) + computa(i2 + 1, n, fila) +
                computa(1, j1 - 1, columna) + computa(j2 + 1, m, columna) -
                computa(i1, i2, fila) - computa(j1, j2, columna)) /
                  2
           << endl;
    }
  }
}
