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 bigint {
  long long a = 0;
  void initiate() { cout << 'i' << endl; }
  void update() { cout << 'u' << endl; }
  void update(long long b) { a += b; }
  long long query() { return a; }
};
template <typename elem_t, typename coord_t, coord_t n_inf, typename ret_t>
class BIT {
  vector<coord_t> positions;
  vector<elem_t> elems;
  bool initiated = false;

 public:
  BIT() { positions.push_back(n_inf); }
  void initiate() {
    if (initiated) {
      for (elem_t &c_elem : elems) {
        c_elem.initiate();
      }
    } else {
      initiated = true;
      sort(positions.begin(), positions.end());
      positions.resize(unique(positions.begin(), positions.end()) -
                       positions.begin());
      elems.resize(positions.size());
    }
  }
  template <typename... loc_form>
  void update(coord_t cord, loc_form... args) {
    if (initiated) {
      int pos = lower_bound(positions.begin(), positions.end(), cord) -
                positions.begin();
      for (; pos < positions.size(); pos += pos & -pos) {
        elems[pos].update(args...);
      }
    } else {
      positions.push_back(cord);
    }
  }
  template <typename... loc_form>
  ret_t query(coord_t cord, loc_form... args) {
    ret_t res = 0;
    int pos = lower_bound(positions.begin(), positions.end(), cord) -
              positions.begin();
    for (; pos > 0; pos -= pos & -pos) {
      res += elems[pos].query(args...);
    }
    return res;
  }
};
struct q {
  int x1, x2, y1, y2;
  int v;
  int type;
};
int n, m, w;
int nr, mr;
q todo[100111];
long long total;
int main() {
  scanf("%d %d %d", &n, &m, &w);
  for (int i = 0; i < w; ++i) {
    scanf("%d %d %d %d %d", &todo[i].type, &todo[i].x1, &todo[i].y1,
          &todo[i].x2, &todo[i].y2);
    if (!todo[i].type) {
      ++todo[i].x2;
      ++todo[i].y2;
      scanf("%d", &todo[i].v);
    } else {
      --todo[i].x1;
      --todo[i].y1;
    }
  }
  BIT<bigint, int, INT_MIN, long long> x, xc, y, yc;
  for (int j = 0; j < w; ++j) {
    x.update(todo[j].x1);
    x.update(todo[j].x2);
    y.update(todo[j].y1);
    y.update(todo[j].y2);
    xc.update(todo[j].x1);
    xc.update(todo[j].x2);
    yc.update(todo[j].y1);
    yc.update(todo[j].y2);
  }
  x.initiate();
  y.initiate();
  xc.initiate();
  yc.initiate();
  for (int i = 0; i < w; ++i) {
    int x1 = todo[i].x1, x2 = todo[i].x2, y1 = todo[i].y1, y2 = todo[i].y2;
    long long v = todo[i].v;
    long long x1r = x1, x2r = x2, y1r = y1, y2r = y2;
    if (todo[i].type) {
      long long res = 0;
      res += xc.query(x2);
      res -= xc.query(x1);
      res += x.query(x2) * x2r;
      res -= x.query(x1) * x1r;
      res += yc.query(y2);
      res -= yc.query(y1);
      res += y.query(y2) * y2r;
      res -= y.query(y1) * y1r;
      res -= total;
      printf("%lld\n", res);
    } else {
      --x1r;
      --x2r;
      --y1r;
      --y2r;
      total += (x2r - x1r) * (y2r - y1r) * v;
      x.update(x1, v * (y2r - y1r));
      x.update(x2, -v * (y2r - y1r));
      xc.update(x1, -v * (y2r - y1r) * x1r);
      xc.update(x2, v * (y2r - y1r) * x2r);
      y.update(y1, v * (x2r - x1r));
      y.update(y2, -v * (x2r - x1r));
      yc.update(y1, -v * (x2r - x1r) * y1r);
      yc.update(y2, v * (x2r - x1r) * y2r);
    }
  }
  return 0;
}
