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;
void _fill_int(int* p, int val, int rep) {
  int i;
  for (i = 0; i < rep; i++) p[i] = val;
}
signed long long GETi() {
  signed long long i;
  scanf("%lld", &i);
  return i;
}
template <class V, int ME>
class BIT_r {
 public:
  V bit[2][1 << ME];
  BIT_r() { clear(); };
  void clear() { memset(bit, 0, sizeof(bit)); };
  void update(int entry, V val0, V val1) {
    entry++;
    while (entry <= 1 << ME)
      bit[0][entry - 1] += val0, bit[1][entry - 1] += val1,
          entry += entry & -entry;
  }
  V total(int entry) {
    int e = entry++;
    V v0 = 0, v1 = 0;
    while (entry > 0)
      v0 += bit[0][entry - 1], v1 += bit[1][entry - 1], entry -= entry & -entry;
    return e * v0 + v1;
  }
  void add(int L, int R, V val) {
    update(L, val, -val * (L - 1));
    update(R + 1, -val, val * R);
  }
};
BIT_r<signed long long, 23> btx, bty;
int R, C, W;
signed long long tot;
void solve() {
  int f, i, j, k, l, x, y;
  signed long long x1, x2, y1, y2, v;
  cin >> R >> C >> W;
  for (i = 0; i < W; i++) {
    cin >> j;
    if (j == 0) {
      cin >> x1 >> y1 >> x2 >> y2 >> v;
      tot += (x2 - x1 + 1) * (y2 - y1 + 1) * v;
      btx.add(x1, x2, v * (y2 - y1 + 1));
      bty.add(y1, y2, v * (x2 - x1 + 1));
    } else {
      cin >> x1 >> y1 >> x2 >> y2;
      signed long long r1 = bty.total(y2) - bty.total(y1 - 1);
      signed long long r2 = btx.total(x2) - btx.total(x1 - 1);
      (void)printf("%lld\n", r1 + r2 - tot);
    }
  }
}
int main(int argc, char** argv) {
  string s;
  if (argc == 1) ios::sync_with_stdio(false);
  for (int i = 1; i < argc; i++) s += argv[i], s += '\n';
  for (int i = s.size() - 1; i >= 0; i--) ungetc(s[i], stdin);
  solve();
  return 0;
}
