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;
inline long long in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline string getStr() {
  char ch[1001];
  scanf("%s", ch);
  return ch;
}
inline char getCh() {
  char ch;
  scanf(" %c", &ch);
  return ch;
}
template <class P, class Q>
inline P smin(P &a, Q b) {
  if (b < a) a = b;
  return a;
}
template <class P, class Q>
inline P smax(P &a, Q b) {
  if (a < b) a = b;
  return a;
}
const long long MOD = 1e9 + 7;
const long long MAX_N = 4e6 + 10;
const long long MAX_LG = 21;
const long long base = 29;
const long long MAX_DIF = 5005;
long long res = 0;
long long fen[MAX_N][4];
inline void add(long long x, long long val, long long tp) {
  for (long long p = x; p < MAX_N; p += p & -p) fen[p][tp] += val;
}
inline long long get(long long x, long long tp) {
  long long res = 0;
  for (long long p = x; p; p -= p & -p) res += fen[p][tp];
  return res;
}
long long cnt;
int32_t main() {
  long long n = in(), m = in(), q = in();
  while (q--) {
    long long tp = in(), x1 = in(), y1 = in(), x2 = in(), y2 = in();
    if (!tp) {
      long long v = in();
      add(x1, 1LL * (y2 - y1 + 1) * v, 0);
      add(x2 + 1, -1LL * (y2 - y1 + 1) * v, 0);
      add(x1, 1LL * v * (x1 - 1) * (y2 - y1 + 1), 1);
      add(x2 + 1, -1LL * v * (y2 - y1 + 1) * x2, 1);
      add(y1, 1LL * (x2 - x1 + 1) * v, 2);
      add(y2 + 1, -1LL * (x2 - x1 + 1) * v, 2);
      add(y1, 1LL * (x2 - x1 + 1) * v * (y1 - 1), 3);
      add(y2 + 1, -1LL * (x2 - x1 + 1) * v * y2, 3);
      cnt += 1LL * v * (x2 - x1 + 1) * (y2 - y1 + 1);
    } else {
      long long row = 1LL * get(x2, 0) * x2 - get(x2, 1) -
                      (1LL * get(x1 - 1, 0) * (x1 - 1) - get(x1 - 1, 1));
      long long column = 1LL * get(y2, 2) * y2 - get(y2, 3) -
                         (1LL * get(y1 - 1, 2) * (y1 - 1) - get(y1 - 1, 3));
      cout << row + column - cnt << "\n";
    }
  }
}
