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<long long> Tx, Ty;
vector<long long> Lx, Ly;
int lx, sizex;
int ly, sizey;
void Stworzx(int n) {
  int p = 1;
  while (p < n) p *= 2;
  lx = p;
  sizex = 2 * p - 1;
  for (int i = 0; i <= sizex; ++i) {
    Tx.push_back(0);
    Lx.push_back(0);
  }
}
void Stworzy(int n) {
  int p = 1;
  while (p < n) p *= 2;
  ly = p;
  sizey = 2 * p - 1;
  for (int i = 0; i <= sizey; ++i) {
    Ty.push_back(0);
    Ly.push_back(0);
  }
}
void Propagujx(int akt, long long L, long long P) {
  Tx[2 * akt] += Lx[akt] * ((P - L + 1) / 2);
  Tx[2 * akt + 1] += Lx[akt] * ((P - L + 1) / 2);
  Lx[2 * akt] += Lx[akt];
  Lx[2 * akt + 1] += Lx[akt];
  Lx[akt] = 0;
}
void Propagujy(int akt, long long L, long long P) {
  Ty[2 * akt] += Ly[akt] * ((P - L + 1) / 2);
  Ty[2 * akt + 1] += Ly[akt] * ((P - L + 1) / 2);
  Ly[2 * akt] += Ly[akt];
  Ly[2 * akt + 1] += Ly[akt];
  Ly[akt] = 0;
}
void Dodajx(int akt, long long L, long long P, long long l, long long p,
            long long x) {
  if (l > p) return;
  if (l == L && p == P) {
    Tx[akt] += (P - L + 1) * x;
    Lx[akt] += x;
  } else {
    Propagujx(akt, L, P);
    long long S = (L + P) / 2;
    Dodajx(2 * akt, L, S, l, min(p, S), x);
    Dodajx(2 * akt + 1, S + 1, P, max(l, S + 1), p, x);
    Tx[akt] = Tx[2 * akt] + Tx[2 * akt + 1];
  }
}
void Dodajy(int akt, long long L, long long P, long long l, long long p,
            long long x) {
  if (l > p) return;
  if (l == L && p == P) {
    Ty[akt] += (P - L + 1) * x;
    Ly[akt] += x;
  } else {
    Propagujy(akt, L, P);
    long long S = (L + P) / 2;
    Dodajy(2 * akt, L, S, l, min(p, S), x);
    Dodajy(2 * akt + 1, S + 1, P, max(l, S + 1), p, x);
    Ty[akt] = Ty[2 * akt] + Ty[2 * akt + 1];
  }
}
long long Sumax(int akt, long long L, long long P, long long l, long long p) {
  if (l > p) return 0;
  if (l == L && p == P) return Tx[akt];
  Propagujx(akt, L, P);
  long long S = (L + P) / 2;
  return Sumax(2 * akt, L, S, l, min(p, S)) +
         Sumax(2 * akt + 1, S + 1, P, max(l, S + 1), p);
}
long long Sumay(int akt, long long L, long long P, long long l, long long p) {
  if (l > p) return 0;
  if (l == L && p == P) return Ty[akt];
  Propagujy(akt, L, P);
  long long S = (L + P) / 2;
  return Sumay(2 * akt, L, S, l, min(p, S)) +
         Sumay(2 * akt + 1, S + 1, P, max(l, S + 1), p);
}
int main() {
  int n, m, w;
  scanf("%d %d %d", &n, &m, &w);
  Stworzx(n);
  Stworzy(m);
  long long wsz = 0;
  for (int i = 0; i < w; ++i) {
    int op, x1, y1, x2, y2;
    scanf("%d %d %d %d %d", &op, &x1, &y1, &x2, &y2);
    --x1;
    --y1;
    --x2;
    --y2;
    if (op == 0) {
      long long v;
      scanf("%I64d", &v);
      Dodajx(1, 0, (sizex + 1) / 2 - 1, x1, x2, (y2 - y1 + 1) * v);
      Dodajy(1, 0, (sizey + 1) / 2 - 1, y1, y2, (x2 - x1 + 1) * v);
      wsz += (x2 - x1 + 1) * (y2 - y1 + 1) * v;
    } else
      printf("%I64d\n", Sumax(1, 0, (sizex + 1) / 2 - 1, x1, x2) +
                            Sumay(1, 0, (sizey + 1) / 2 - 1, y1, y2) - wsz);
  }
  return 0;
}
