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;
const int N = 100100;
long long T[(1 << 23) + 10][4];
int off = 1, lo, hi;
int idx;
long long VAL;
int n, m, w;
int VEL[5];
inline void refresh(int a, int b, int node) {
  b = min(VEL[idx] - 1, b);
  if (T[node][idx + 1]) {
    T[node][idx] += (long long)(b - a + 1) * T[node][idx + 1];
    if (a != b) {
      T[node * 2][idx + 1] += T[node][idx + 1];
      T[node * 2 + 1][idx + 1] += T[node][idx + 1];
    }
    T[node][idx + 1] = 0;
  }
}
void update(int a, int b, int node) {
  refresh(a, b, node);
  if (a > hi || b < lo) return;
  if (a >= lo && b <= hi) {
    T[node][idx + 1] += VAL;
    refresh(a, b, node);
    return;
  }
  update(a, (a + b) / 2, node * 2);
  update((a + b) / 2 + 1, b, node * 2 + 1);
  T[node][idx] = T[node * 2][idx] + T[node * 2 + 1][idx];
}
long long query(int a, int b, int node) {
  refresh(a, b, node);
  if (a > hi || b < lo) return 0;
  if (a >= lo && b <= hi) return T[node][idx];
  return query(a, (a + b) / 2, node * 2) +
         query((a + b) / 2 + 1, b, node * 2 + 1);
}
int main(void) {
  scanf("%d %d %d", &n, &m, &w);
  VEL[0] = n;
  VEL[2] = m;
  while (off < max(n, m)) off <<= 1;
  for (; w; w--) {
    int ty, x1, y1, x2, y2;
    scanf("%d %d %d %d %d", &ty, &x1, &y1, &x2, &y2);
    x1--;
    y1--;
    x2--;
    y2--;
    if (ty == 0) {
      int v;
      scanf("%d", &v);
      VAL = (long long)(y2 - y1 + 1) * v;
      lo = x1;
      hi = x2;
      idx = 0;
      update(0, off - 1, 1);
      VAL = (long long)(x2 - x1 + 1) * v;
      lo = y1;
      hi = y2;
      idx = 2;
      update(0, off - 1, 1);
    } else {
      lo = 0, hi = x1 - 1;
      idx = 0;
      long long ans = query(0, off - 1, 1);
      lo = x2 + 1, hi = n - 1;
      long long ans2 = query(0, off - 1, 1);
      idx = 2;
      lo = y1, hi = y2;
      long long ans3 = query(0, off - 1, 1);
      cout << ans3 - ans2 - ans << endl;
    }
  }
  return 0;
}
