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 maxn = 4e6;
long long b1[maxn], b2[maxn], b3[maxn], b4[maxn], tot;
int n, m, q;
inline int Lowbit(int i) { return i & -i; }
void Add(long long *b1, int n, int i, long long x) {
  for (; i < n; i += Lowbit(i + 1)) {
    b1[i] += x;
  }
}
long long Sum(const long long *b1, int i) {
  long long s = 0;
  for (--i; i >= 0; i -= Lowbit(i + 1)) {
    s += b1[i];
  }
  return s;
}
void Add(long long *b1, long long *b2, int n, int l, int r, long long x) {
  Add(b2, n, l, x);
  Add(b2, n, r, -x);
  Add(b1, n, l, -l * x);
  Add(b1, n, r, r * x);
}
long long Sum(const long long *b1, const long long *b2, int l, int r) {
  long long s = Sum(b1, r) - Sum(b1, l);
  s += Sum(b2, r) * r;
  s -= Sum(b2, l) * l;
  return s;
}
int main(void) {
  scanf("%d%d%d", &n, &m, &q);
  while (q--) {
    int t, x1, y1, x2, y2;
    scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2);
    --x1;
    --y1;
    if (t == 0) {
      int w;
      scanf("%d", &w);
      tot += (long long)(x2 - x1) * (y2 - y1) * w;
      Add(b1, b2, n, x1, x2, (long long)(y2 - y1) * w);
      Add(b3, b4, m, y1, y2, (long long)(x2 - x1) * w);
    } else {
      long long ans = -tot;
      ans += Sum(b1, b2, x1, x2);
      ans += Sum(b3, b4, y1, y2);
      printf("%lld\n", ans);
    }
  }
  return 0;
}
