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;
struct Debug {
  Debug() {}
  template <class T>
  inline Debug operator<<(T a) {
    return *this;
  }
} debug;
const int MAX = 4 * 1000 * 1000 + 5;
class FenwickTreeLazyUpdate {
 private:
  int len;
  long long *a, *b;
  inline void add_a(const int &pos, const long long &val) {
    for (int i = pos; i <= len; i |= i - 1, ++i) a[i] += val;
  }
  inline void add_b(const int &pos, const long long &val) {
    for (int i = pos; i <= len; i |= i - 1, ++i) b[i] += val;
  }
  inline long long get_a(const int &pos) {
    long long res = 0;
    for (int i = pos; i >= 1; i &= i - 1) res += a[i];
    return res;
  }
  inline long long get_b(const int &pos) {
    long long res = 0;
    for (int i = pos; i >= 1; i &= i - 1) res += b[i];
    return res;
  }
  inline long long sum(int pos) { return get_a(pos) * pos - get_b(pos); }

 public:
  FenwickTreeLazyUpdate(const int &size) {
    len = size;
    a = new long long[size];
    b = new long long[size];
  }
  inline void update(const int &left, const int &right, const long long &val) {
    add_a(left, val);
    add_a(right + 1, -val);
    add_b(left, (left - 1) * val);
    add_b(right + 1, -right * val);
  }
  inline long long sum(const int &left, const int &right) {
    return sum(right) - sum(left - 1);
  }
} line(MAX), colm(MAX);
int n, m, w;
int main() {
  cin >> n >> m >> w;
  for (; w--;) {
    int type;
    long long val;
    pair<int, int> x, y;
    cin >> type;
    cin >> x.first >> y.first >> x.second >> y.second;
    if (type == 0)
      cin >> val,
          line.update(x.first, x.second, val * (y.second - y.first + 1)),
          colm.update(y.first, y.second, val * (x.second - x.first + 1));
    else
      cout << line.sum(x.first, x.second) - colm.sum(1, y.first - 1) -
                  colm.sum(y.second + 1, m)
           << '\n';
  }
}
