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 STree {
  vector<long long> st, lazy;
  int n;
  STree(int n) : st(4 * n + 5, 0), lazy(4 * n + 5, 0), n(n) {}
  void push(int k, int s, int e) {
    if (!lazy[k]) return;
    st[k] += (e - s) * lazy[k];
    if (s + 1 < e) {
      lazy[2 * k] += lazy[k];
      lazy[2 * k + 1] += lazy[k];
    }
    lazy[k] = 0;
  }
  void upd(int k, int s, int e, int a, int b, long long v) {
    push(k, s, e);
    if (s >= b || e <= a) return;
    if (s >= a && e <= b) {
      lazy[k] += v;
      push(k, s, e);
      return;
    }
    int m = (s + e) / 2;
    upd(2 * k, s, m, a, b, v);
    upd(2 * k + 1, m, e, a, b, v);
    st[k] = st[2 * k] + st[2 * k + 1];
  }
  long long query(int k, int s, int e, int a, int b) {
    if (s >= b || e <= a) return 0;
    push(k, s, e);
    if (s >= a && e <= b) return st[k];
    int m = (s + e) / 2;
    return query(2 * k, s, m, a, b) + query(2 * k + 1, m, e, a, b);
  }
  void upd(int a, int b, long long v) { upd(1, 0, n, a, b, v); }
  long long query(int a, int b) { return query(1, 0, n, a, b); }
};
int n, m, q;
int main() {
  scanf("%d%d%d", &n, &m, &q);
  STree s0(n), s1(m);
  while (q--) {
    int t, x0, x1, y0, y1;
    scanf("%d%d%d%d%d", &t, &x0, &y0, &x1, &y1);
    x0--;
    y0--;
    if (t == 0) {
      int v;
      scanf("%d", &v);
      s0.upd(x0, x1, 1LL * (y1 - y0) * v);
      s1.upd(y0, y1, 1LL * (x1 - x0) * v);
    } else
      printf("%lld\n", s0.query(x0, x1) + s1.query(y0, y1) - s0.query(0, n));
  }
  return 0;
}
