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 N = 4e6 + 10;
int n, m, w;
template <class T>
struct Tree {
  long long c[2][N];
  int maxn;
  void init(int x) {
    maxn = x + 10;
    memset(c, 0, sizeof c);
  }
  inline int lowbit(int x) { return x & -x; }
  long long sum(long long *b, int x) {
    long long ans = 0;
    if (x == 0) ans = b[0];
    while (x) ans += b[x], x -= lowbit(x);
    return ans;
  }
  void change(long long *b, int x, long long value) {
    if (x == 0) b[x] += value, x++;
    while (x <= maxn) b[x] += value, x += lowbit(x);
  }
  long long get_pre(int r) { return sum(c[0], r) * r + sum(c[1], r); }
  void add(int l, int r, long long value) {
    change(c[0], l, value);
    change(c[0], r + 1, -value);
    change(c[1], l, value * (-l + 1));
    change(c[1], r + 1, value * r);
  }
  long long get(int l, int r) { return get_pre(r) - get_pre(l - 1); }
};
Tree<long long> x, y;
int main() {
  scanf("%d%d%d", &n, &m, &w);
  x.init(n);
  y.init(m);
  int tmp;
  long long all = 0;
  while (w--) {
    scanf("%d", &tmp);
    int x1, x2, y1, y2, v;
    if (tmp == 0) {
      scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &v);
      all += v * (x2 - x1 + 1) * (y2 - y1 + 1);
      x.add(x1, x2, v * (y2 - y1 + 1));
      y.add(y1, y2, v * (x2 - x1 + 1));
    }
    if (tmp == 1) {
      scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
      printf("%I64d\n", y.get(1, y2) - y.get(1, y1 - 1) -
                            (all - x.get(1, x2) + x.get(1, x1 - 1)));
    }
  }
  return 0;
}
