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;
long long Sx[4000011 << 2], Sy[4000011 << 2];
int xa, n, tip, m, w, ya, xb, yb, x;
long long sum;
void updx(int st, int dr, int po, int l, int r, long long val) {
  if (st >= l && dr <= r) {
    Sx[po] += val * (dr - st + 1);
    return;
  }
  int mij = (st + dr) >> 1;
  if (Sx[po] != Sx[po << 1] + Sx[(po << 1) + 1]) {
    long long dif = Sx[po] - (Sx[po << 1] + Sx[(po << 1) + 1]);
    dif /= (dr - st + 1);
    Sx[po << 1] += dif * (mij - st + 1);
    Sx[(po << 1) + 1] += dif * (dr - mij);
  }
  if (l <= mij) updx(st, mij, po << 1, l, r, val);
  if (r > mij) updx(mij + 1, dr, (po << 1) + 1, l, r, val);
  Sx[po] = Sx[po << 1] + Sx[(po << 1) + 1];
}
void finx(int st, int dr, int po, int l, int r, int sg) {
  if (st >= l && dr <= r) {
    sum += Sx[po] * sg;
    return;
  }
  int mij = (st + dr) >> 1;
  if (Sx[po] != Sx[po << 1] + Sx[(po << 1) + 1]) {
    long long dif = Sx[po] - (Sx[po << 1] + Sx[(po << 1) + 1]);
    dif /= (dr - st + 1);
    Sx[po << 1] += dif * (mij - st + 1);
    Sx[(po << 1) + 1] += dif * (dr - mij);
  }
  if (l <= mij) finx(st, mij, po << 1, l, r, sg);
  if (r > mij) finx(mij + 1, dr, (po << 1) + 1, l, r, sg);
}
void updy(int st, int dr, int po, int l, int r, long long val) {
  if (st >= l && dr <= r) {
    Sy[po] += val * (dr - st + 1);
    return;
  }
  int mij = (st + dr) >> 1;
  if (Sy[po] != Sy[po << 1] + Sy[(po << 1) + 1]) {
    long long dif = Sy[po] - (Sy[po << 1] + Sy[(po << 1) + 1]);
    dif /= (dr - st + 1);
    Sy[po << 1] += dif * (mij - st + 1);
    Sy[(po << 1) + 1] += dif * (dr - mij);
  }
  if (l <= mij) updy(st, mij, po << 1, l, r, val);
  if (r > mij) updy(mij + 1, dr, (po << 1) + 1, l, r, val);
  Sy[po] = Sy[po << 1] + Sy[(po << 1) + 1];
}
void finy(int st, int dr, int po, int l, int r, int sg) {
  if (st >= l && dr <= r) {
    sum += Sy[po] * sg;
    return;
  }
  int mij = (st + dr) >> 1;
  if (Sy[po] != Sy[po << 1] + Sy[(po << 1) + 1]) {
    long long dif = Sy[po] - (Sy[po << 1] + Sy[(po << 1) + 1]);
    dif /= (dr - st + 1);
    Sy[po << 1] += dif * (mij - st + 1);
    Sy[(po << 1) + 1] += dif * (dr - mij);
  }
  if (l <= mij) finy(st, mij, po << 1, l, r, sg);
  if (r > mij) finy(mij + 1, dr, (po << 1) + 1, l, r, sg);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> w;
  int dim = 4000001;
  for (int i = 1; i <= w; ++i) {
    cin >> tip;
    if (!tip) {
      cin >> xa >> ya >> xb >> yb >> x;
      updx(1, dim, 1, xa, xb, 1LL * x * (yb - ya + 1));
      updy(1, dim, 1, ya, yb, 1LL * x * (xb - xa + 1));
    } else {
      cin >> xa >> ya >> xb >> yb;
      sum = 0;
      finx(1, dim, 1, xa, xb, 1);
      finy(1, dim, 1, ya, yb, 1);
      finx(1, dim, 1, 1, xa - 1, -1);
      finx(1, dim, 1, xb + 1, dim, -1);
      finy(1, dim, 1, 1, ya - 1, -1);
      finy(1, dim, 1, yb + 1, dim, -1);
      sum /= 2;
      cout << sum << "\n";
    }
  }
  return 0;
}
