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;
double PI = acos(-1);
double EPS = 1e-7;
int INF = 1000000000;
int MOD = 1000000007;
int MAXINT = 2147483647;
long long INFLL = 1000000000000000000LL;
long long MAXLL = 9223372036854775807LL;
int mx[8] = {-1, 1, 0, 0, -1, -1, 1, 1};
int my[8] = {0, 0, -1, 1, -1, 1, -1, 1};
long long bit[2][2][4000005];
void update(long long bit[], int x, long long v) {
  for (int i = x; i <= 4000000; i += (i & -i)) {
    bit[i] += v;
  }
}
long long query(long long bit[], int x) {
  long long res = 0;
  for (int i = x; i >= 1; i -= (i & -i)) {
    res += bit[i];
  }
  return res;
}
long long query(int t, int x) {
  return (long long)query(bit[t][0], x) * (long long)x + query(bit[t][1], x);
}
long long query(int t, int l, int r) { return query(t, r) - query(t, l - 1); }
void update(int t, int l, int r, long long v) {
  update(bit[t][0], l, (long long)v);
  update(bit[t][0], r + 1, (long long)-v);
  update(bit[t][1], l, (long long)-v * (long long)(l - 1));
  update(bit[t][1], r + 1, (long long)v * (long long)r);
}
int main() {
  int n, m, w;
  scanf("%d%d%d", &n, &m, &w);
  long long sum = 0;
  for (int(a) = (1); (a) <= (w); (a)++) {
    int t, x1, x2, y1, y2, v;
    scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2);
    if (t == 0) {
      scanf("%d", &v);
      update(0, x1, x2, (long long)(y2 - y1 + 1) * (long long)v);
      update(1, y1, y2, (long long)(x2 - x1 + 1) * (long long)v);
      sum += (long long)(y2 - y1 + 1) * (x2 - x1 + 1) * v;
    } else {
      printf("%I64d\n", -sum + query(0, x1, x2) + query(1, y1, y2));
    }
  }
}
