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;
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, Q;
long long mx;
struct Tree {
  long long del[5000005], dei[5000005];
  int n;
  void Clear(int u) {
    memset(del, 0, sizeof(del));
    memset(dei, 0, sizeof(dei));
    n = u;
  }
  int lowbit(int x) { return x & (-x); }
  void Add(int x, long long y) {
    for (int i = x; i; i -= lowbit(i)) del[i] += y;
    for (int i = x; i <= n; i += lowbit(i)) dei[i] += x * y;
  }
  long long Sum(int x) {
    long long ans = 0, ans2 = 0;
    for (int i = x; i <= n; i += lowbit(i)) ans += del[i];
    for (int i = x - 1; i; i -= lowbit(i)) ans2 += dei[i];
    return ans * x + ans2;
  }
  long long Query(int x, int y) { return Sum(y) - Sum(x - 1); }
} X, Y;
int main() {
  n = read();
  m = read();
  Q = read();
  mx = 0;
  X.Clear(m);
  Y.Clear(n);
  while (Q--) {
    int opt = read(), x1 = read(), y1 = read(), x2 = read(), y2 = read();
    if (opt == 0) {
      long long v = read(), addx, addy;
      addx = v * (y2 - y1 + 1);
      addy = v * (x2 - x1 + 1);
      Y.Add(x2, addx);
      if (x1 - 1) Y.Add(x1 - 1, -addx);
      X.Add(y2, addy);
      if (y1 - 1) X.Add(y1 - 1, -addy);
      mx += (y2 - y1 + 1) * (x2 - x1 + 1) * v;
    }
    if (opt == 1) {
      long long ans = -mx;
      ans += X.Query(y1, y2);
      ans += Y.Query(x1, x2);
      printf("%I64d\n", ans);
    }
  }
  return 0;
}
