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 Fenwick {
  long long *m, *mt, N;

 public:
  Fenwick(long n);
  Fenwick(long long a[], long n);
  void add_range(long l, long long d);
  void add_range(long l, long r, long long d);
  long long sum(long r);
  long long sum(long l, long r);
};
Fenwick::Fenwick(long n) {
  N = n;
  m = new long long[N];
  mt = new long long[N];
  for (int i = 0; i < n; i++) m[i] = mt[i] = 0;
}
void Fenwick::add_range(long r, long long d) {
  if (r < 0) return;
  for (int i = r; i >= 0; i = (i & (i + 1)) - 1) mt[i] += d;
  for (int i = r | (r + 1); i < N; i |= i + 1)
    m[i] += d * (r - (i & (i + 1)) + 1);
}
void Fenwick::add_range(long l, long r, long long d) {
  add_range(r, d);
  add_range(l - 1, -d);
}
long long Fenwick::sum(long r) {
  if (r < 0) return 0;
  long long res = 0;
  for (int i = r; i >= 0; i = (i & (i + 1)) - 1)
    res += m[i] + mt[i] * (i - (i & (i + 1)) + 1);
  for (int i = r | (r + 1); i < N; i |= i + 1)
    res += mt[i] * (r - (i & (i + 1)) + 1);
  return res;
}
long long Fenwick::sum(long l, long r) { return sum(r) - sum(l - 1); }
long n, m, w;
long long tp, x1, aasdfasdfasdf, x2, y2;
long long v, ttl;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> w;
  Fenwick f1(n + 100), f2(m + 100);
  for (; w--;) {
    cin >> tp;
    if (tp == 0) {
      cin >> x1 >> aasdfasdfasdf >> x2 >> y2 >> v;
      ttl += (x2 - x1 + 1) * (y2 - aasdfasdfasdf + 1) * v;
      f1.add_range(x1, x2, (y2 - aasdfasdfasdf + 1) * v);
      f2.add_range(aasdfasdfasdf, y2, (x2 - x1 + 1) * v);
    } else {
      cin >> x1 >> aasdfasdfasdf >> x2 >> y2;
      long long res = -ttl;
      res += f1.sum(x1, x2);
      res += f2.sum(aasdfasdfasdf, y2);
      cout << res << endl;
    }
  }
  cin.get();
  cin.get();
  return 0;
}
