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;
template <typename Abel>
struct BIT {
  vector<Abel> dat[2];
  BIT(int n) {
    for (int id : {0, 1}) dat[id].assign(n + 1, Abel(0));
  }
  inline void sub_add(int id, int a, Abel x) {
    for (int i = a; i < (int)dat[id].size(); i += i & -i) dat[id][i] += x;
  }
  inline void add(int a, int b, Abel x) {
    sub_add(0, a, -x * (a - 1));
    sub_add(1, a, x);
    sub_add(0, b, x * (b - 1));
    sub_add(1, b, -x);
  }
  inline Abel sub_sum(int id, int a) {
    Abel res = Abel(0);
    for (int i = a; i > 0; i -= i & -i) res += dat[id][i];
    return res;
  }
  inline Abel sum(int a, int b) {
    Abel res = sub_sum(0, b - 1) + sub_sum(1, b - 1) * (b - 1);
    res -= sub_sum(0, a - 1) + sub_sum(1, a - 1) * (a - 1);
    return res;
  }
};
int main() {
  int h, w;
  cin >> h >> w;
  int q;
  cin >> q;
  BIT<long long> bit_h(h), bit_w(w);
  long long tot = 0;
  while (q--) {
    int op;
    cin >> op;
    int si, sj, ti, tj;
    cin >> si >> sj >> ti >> tj;
    ti++, tj++;
    if (op == 0) {
      long long v;
      cin >> v;
      tot += v * (ti - si) * (tj - sj);
      bit_h.add(si, ti, v * (tj - sj));
      bit_w.add(sj, tj, v * (ti - si));
    }
    if (op == 1) {
      cout << bit_h.sum(si, ti) + bit_w.sum(sj, tj) - tot << endl;
    }
  }
  return 0;
}
