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 n, m, w;
long long rowa[4000010], rowb[4000010], cola[4000010], colb[4000010];
void upd(long long *treea, long long *treeb, int pos, long long a,
         long long b) {
  for (; pos < 4000010; pos |= pos + 1) {
    treea[pos] += a;
    treeb[pos] += b;
  }
}
void update(long long *treea, long long *treeb, int l, int r, long long val) {
  upd(treea, treeb, l, val, -val * (l - 1));
  upd(treea, treeb, r, -val, val * r);
}
long long get(long long *treea, long long *treeb, int pos) {
  long long x = pos, a = 0, b = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) {
    a += treea[pos];
    b += treeb[pos];
  }
  return a * x + b;
}
long long query(long long *treea, long long *treeb, int l, int r) {
  return get(treea, treeb, r) - get(treea, treeb, l - 1);
}
int main() {
  cin >> n >> m >> w;
  while (w--) {
    int op, l, r, u, d;
    cin >> op >> u >> l >> d >> r;
    if (op == 0) {
      long long v;
      cin >> v;
      update(rowa, rowb, u, d, v * (r - l + 1));
      update(cola, colb, l, r, v * (d - u + 1));
    } else {
      long long ans = query(cola, colb, l, r);
      ans -= query(rowa, rowb, 1, u - 1);
      ans -= query(rowa, rowb, d + 1, n);
      cout << ans << '\n';
    }
  }
}
