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;
class BIT {
 private:
  int n;
  vector<long long> mul, add;
  void updatepoint(int x, long long m, long long a) {
    for (; x <= n; x += x & (-x)) {
      mul[x] += m;
      add[x] += a;
    }
  }
  long long getpoint(int x) {
    long long m = 0LL;
    long long a = 0LL;
    long long f = x;
    for (; x >= 1; x = x & (x - 1)) {
      m += mul[x];
      a += add[x];
    }
    return (m * f + a);
  }

 public:
  BIT() { n = 0; }
  BIT(int n) {
    this->n = n;
    mul = vector<long long>(n + 7, 0LL);
    add = vector<long long>(n + 7, 0LL);
  }
  void update(int l, int r, long long x) {
    updatepoint(l, x, -x * (l - 1));
    updatepoint(r, -x, x * r);
  }
  long long getsum(int l, int r) { return (getpoint(r) - getpoint(l - 1)); }
};
BIT row, col;
int m, n, q;
long long s;
void add(int x1, int y1, int x2, int y2, int v) {
  row.update(x1, x2, 1LL * (y2 - y1 + 1) * v);
  col.update(y1, y2, 1LL * (x2 - x1 + 1) * v);
}
long long answer(int x1, int y1, int x2, int y2) {
  return (row.getsum(x1, x2) + col.getsum(y1, y2) - s);
}
void process(void) {
  cin >> m >> n >> q;
  row = BIT(m);
  col = BIT(n);
  for (int i = 0; i < (q); i = i + 1) {
    int t, x1, y1, x2, y2;
    cin >> t >> x1 >> y1 >> x2 >> y2;
    if (t < 1) {
      int v;
      cin >> v;
      add(x1, y1, x2, y2, v);
      s += 1LL * (x2 - x1 + 1) * (y2 - y1 + 1) * v;
    } else
      cout << answer(x1, y1, x2, y2) << "\n";
  }
}
int main(void) {
  ios::sync_with_stdio(false);
  process();
  return 0;
}
