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 nmax = 4e6 + 5;
struct seg {
  long long sum, lazy;
  seg *lc, *rc;
  int L, R, M;
  seg(int _L, int _R) : L(_L), R(_R) {
    sum = lazy = 0;
    lc = rc = NULL;
    M = (L + R) / 2;
  };
  ~seg() {
    delete lc;
    delete rc;
  }
  void upd() {
    lazy_down();
    sum = 0;
    if (lc != NULL) sum += lc->sum;
    if (rc != NULL) sum += rc->sum;
  }
  void make_left() {
    if (lc == NULL) lc = new seg(L, M);
  }
  void make_right() {
    if (rc == NULL) rc = new seg(M + 1, R);
  }
  void lazy_down() {
    if (lazy == 0 || L == R) return;
    make_left(), make_right();
    lc->sum += lazy * (M - L + 1);
    rc->sum += lazy * (R - M);
    lc->lazy += lazy;
    rc->lazy += lazy;
    lazy = 0;
  }
  long long get_sum(int l, int r) {
    lazy_down();
    if (l == L && r == R) {
      return sum;
    }
    long long ans = 0;
    if (l <= M) {
      make_left();
      ans += lc->get_sum(l, min(M, r));
    }
    if (r > M) {
      make_right();
      ans += rc->get_sum(max(M + 1, l), r);
    }
    return ans;
  }
  void add(int l, int r, int val) {
    lazy_down();
    if (l == L && r == R) {
      sum += (R - L + 1) * val;
      lazy += val;
      return;
    }
    if (l <= M) {
      make_left();
      lc->add(l, min(M, r), val);
    }
    if (r > M) {
      make_right();
      rc->add(max(M + 1, l), r, val);
    }
    upd();
  }
};
int n, m, w;
long long total;
int main() {
  scanf("%d %d %d", &n, &m, &w);
  seg x(1, n), y(1, m);
  for (int i = 0; i < w; ++i) {
    int type, x1, y1, x2, y2, val;
    scanf("%d %d %d %d %d", &type, &x1, &y1, &x2, &y2);
    if (type == 0) {
      scanf("%d", &val);
      x.add(x1, x2, val * (y2 - y1 + 1));
      y.add(y1, y2, val * (x2 - x1 + 1));
      total += (x2 - x1 + 1) * (y2 - y1 + 1) * val;
    } else {
      long long ans = 0;
      ans += x.get_sum(x1, x2);
      ans += y.get_sum(y1, y2);
      ans -= total;
      printf("%lld\n", ans);
    }
  }
}
