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;
vector<pair<long long, long long>> segment[2];
int n, m, w, t1, t2;
long long tot;
long long update(int a, int b, int tag, long long val, int curr, int l, int r) {
  if (b < l || a > r) {
    return 0;
  }
  if (a <= l && b >= r) {
    segment[tag][curr].first += (r - l + 1) * val;
    segment[tag][curr].second += val;
    return segment[tag][curr].first;
  }
  int m = (l + r) / 2;
  segment[tag][2 * curr].first += (r - m) * segment[tag][curr].second,
      segment[tag][2 * curr + 1].first += (r - m) * segment[tag][curr].second;
  segment[tag][2 * curr].second += segment[tag][curr].second,
      segment[tag][2 * curr + 1].second += segment[tag][curr].second;
  segment[tag][curr].second = 0;
  long long ret = update(a, b, tag, val, 2 * curr, l, m) +
                  update(a, b, tag, val, 2 * curr + 1, m + 1, r);
  segment[tag][curr].first =
      segment[tag][2 * curr].first + segment[tag][2 * curr + 1].first;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
  cin >> n >> m >> w;
  t1 = pow(2, ceil(log2(n))), t2 = pow(2, ceil(log2(m)));
  segment[0].resize(2 * t1, {0, 0});
  segment[1].resize(2 * t2, {0, 0});
  while (w--) {
    int a, x1, y1, x2, y2;
    cin >> a >> x1 >> y1 >> x2 >> y2;
    if (a == 0) {
      long long v;
      cin >> v;
      tot += (y2 - y1 + 1) * (x2 - x1 + 1) * v;
      update(x1, x2, 0, v * (y2 - y1 + 1), 1, 1, t1);
      update(y1, y2, 1, v * (x2 - x1 + 1), 1, 1, t2);
    } else {
      cout << -tot + update(x1, x2, 0, 0, 1, 1, t1) +
                  update(y1, y2, 1, 0, 1, 1, t2)
           << '\n';
    }
  }
}
