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;
class BIT {
 private:
  vector<long long int> val;
  int n;
  long long int calc(const long long int a, const long long int b) {
    return a + b;
  }

 public:
  BIT() {}
  BIT(int size) {
    n = 1;
    while (n < size) n <<= 1;
    val = vector<long long int>(2 * n, 0LL);
  }
  void add(int x, long long int a) {
    x += n - 1;
    val[x] += a;
    while (x > 0) {
      x = (x - 1) / 2;
      val[x] = calc(val[x * 2 + 1], val[x * 2 + 2]);
    }
  }
  long long int query(int a, int b, int k = 0, int l = 0, int r = -1) {
    if (r == -1) r = n;
    if (r <= a || b <= l) return 0LL;
    if (a <= l && r <= b) {
      return val[k];
    } else {
      return calc(query(a, b, k * 2 + 1, l, (l + r) / 2),
                  query(a, b, k * 2 + 2, (l + r) / 2, r));
    }
  }
};
class Seg {
  BIT dif, pre;

 public:
  Seg(int n) {
    dif = BIT(n);
    pre = BIT(n);
  }
  void add(int s, int t, long long int v) {
    dif.add(s, v);
    dif.add(t, -v);
    pre.add(s, v * s);
    pre.add(t, -v * t);
  }
  long long int sum(int s, int t) {
    if (s > 0) return sum(0, t) - sum(0, s);
    return dif.query(0, t) * t - pre.query(0, t);
  }
};
int main() {
  int n, m, w;
  scanf("%d%d%d", &n, &m, &w);
  Seg segX(n + 1), segY(m + 1);
  long long int sum = 0;
  for (int i = 0; i < (int)(w); i++) {
    int a, x1, x2, y1, y2;
    long long int v;
    scanf("%d%d%d%d%d", &a, &x1, &y1, &x2, &y2);
    if (a) {
      printf("%I64d\n", segX.sum(x1 - 1, x2) + segY.sum(y1 - 1, y2) - sum);
    } else {
      scanf("%I64d", &v);
      segX.add(x1 - 1, x2, v * (y2 - y1 + 1));
      segY.add(y1 - 1, y2, v * (x2 - x1 + 1));
      sum += v * (x2 - x1 + 1) * (y2 - y1 + 1);
    }
  }
  return 0;
}
