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>
const int N = 4e6 + 10;
const int Inf = 0x3f3f3f3f;
using namespace std;
int Read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  return x * f;
}
const int V = 4000009;
long long nm[2], tr[2][2][V];
int lowbit(int x) { return x & -x; }
void Add(int idx1, int idx2, long long val) {
  for (int x = idx2 + 1; x <= nm[idx1]; x += lowbit(x))
    tr[idx1][0][x] += val * idx2, tr[idx1][1][x] += val;
}
long long Query(int idx1, int idx2) {
  long long res = 0;
  for (int x = ++idx2; x > 0; x -= lowbit(x))
    res += tr[idx1][1][x] * idx2 - tr[idx1][0][x];
  return res;
}
int main() {
  long long w, t, x1, y1, x2, y2, v;
  nm[0] = Read();
  nm[1] = Read();
  w = Read();
  while (w--) {
    t = Read();
    x1 = Read();
    y1 = Read();
    x2 = Read();
    y2 = Read();
    --x1, --y1;
    if (!t) {
      scanf("%I64d", &v);
      Add(0, x1, (y2 - y1) * v);
      Add(0, x2, (y1 - y2) * v);
      Add(1, y1, (x2 - x1) * v);
      Add(1, y2, (x1 - x2) * v);
    } else {
      --x2, --y2;
      long long part1 = Query(0, x2) - Query(0, x1 - 1);
      long long part2 = Query(1, y1 - 1);
      long long part3 = Query(1, nm[1] - 1) - Query(1, y2);
      printf("%I64d\n", part1 - part2 - part3);
    }
  }
  return 0;
}
void fastInOut() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
}
