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;
int n, m, q;
long long h[1 << 23], w[1 << 23];
long long ht[1 << 23], wt[1 << 23];
void updateh(int nd, int l, int r, int lt, int rt, long long v) {
  if (l > rt || r < lt) return;
  if (l == r && l == lt) {
    h[nd] += v * (r - l + 1);
    return;
  }
  if (l >= lt && r <= rt) {
    ht[nd] += v * (r - l + 1);
    return;
  }
  int k = (l + r) >> 1;
  long long tmp = ht[nd] / (r - l + 1);
  updateh(2 * nd, l, k, l, r, tmp);
  updateh(2 * nd + 1, k + 1, r, l, r, tmp);
  ht[nd] = 0;
  updateh(2 * nd, l, k, lt, rt, v);
  updateh(2 * nd + 1, k + 1, r, lt, rt, v);
  h[nd] = h[2 * nd] + h[2 * nd + 1] + ht[2 * nd] + ht[2 * nd + 1];
}
void updatev(int nd, int l, int r, int lt, int rt, long long v) {
  if (l > rt || r < lt) return;
  if (l == r && l == lt) {
    w[nd] += v * (r - l + 1);
    return;
  }
  if (l >= lt && r <= rt) {
    wt[nd] += v * (r - l + 1);
    return;
  }
  int k = (l + r) >> 1;
  long long tmp = wt[nd] / (r - l + 1);
  updatev(2 * nd, l, k, l, r, tmp);
  updatev(2 * nd + 1, k + 1, r, l, r, tmp);
  wt[nd] = 0;
  updatev(2 * nd, l, k, lt, rt, v);
  updatev(2 * nd + 1, k + 1, r, lt, rt, v);
  w[nd] = w[2 * nd] + w[2 * nd + 1] + wt[2 * nd] + wt[2 * nd + 1];
}
long long queryh(int nd, int l, int r, int lt, int rt) {
  if (l > rt || r < lt) return 0;
  if (l >= lt && r <= rt) {
    return ht[nd] + h[nd];
  }
  int k = (l + r) >> 1;
  if (ht[nd]) {
    long long tmp = ht[nd] / (r - l + 1);
    updateh(2 * nd, l, k, l, r, tmp);
    updateh(2 * nd + 1, k + 1, r, l, r, tmp);
    ht[nd] = 0;
    h[nd] = h[2 * nd] + h[2 * nd + 1] + ht[2 * nd] + ht[2 * nd + 1];
  }
  long long res =
      queryh(2 * nd, l, k, lt, rt) + queryh(2 * nd + 1, k + 1, r, lt, rt);
  return res;
}
long long queryv(int nd, int l, int r, int lt, int rt) {
  if (l > rt || r < lt) return 0;
  if (l >= lt && r <= rt) {
    return wt[nd] + w[nd];
  }
  int k = (l + r) >> 1;
  if (wt[nd]) {
    long long tmp = wt[nd] / (r - l + 1);
    updatev(2 * nd, l, k, l, r, tmp);
    updatev(2 * nd + 1, k + 1, r, l, r, tmp);
    wt[nd] = 0;
    w[nd] = w[2 * nd] + w[2 * nd + 1] + wt[2 * nd] + wt[2 * nd + 1];
  }
  long long res =
      queryv(2 * nd, l, k, lt, rt) + queryv(2 * nd + 1, k + 1, r, lt, rt);
  return res;
}
void printh(int nd, int l, int r) {
  cout << nd << " " << l << " " << r << " " << h[nd] << " " << ht[nd] << endl;
  if (l == r) return;
  printh(2 * nd, l, (l + r) >> 1);
  printh(2 * nd + 1, ((l + r) >> 1) + 1, r);
}
void printv(int nd, int l, int r) {
  cout << nd << " " << l << " " << r << " " << w[nd] << " " << wt[nd] << endl;
  if (l == r) return;
  printv(2 * nd, l, (l + r) >> 1);
  printv(2 * nd + 1, ((l + r) >> 1) + 1, r);
}
int main() {
  cin >> n >> m >> q;
  long long tot = 0;
  for (int i = 0; i < q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 0) {
      int x1, y1, x2, y2, val;
      scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &val);
      updateh(1, 1, m, y1, y2, (long long)val * (x2 - x1 + 1));
      updatev(1, 1, n, x1, x2, (long long)val * (y2 - y1 + 1));
      tot += (long long)val * (x2 - x1 + 1) * (y2 - y1 + 1);
    }
    if (t == 1) {
      int x1, y1, x2, y2;
      scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
      long long ans = queryh(1, 1, m, y1, y2) + queryv(1, 1, n, x1, x2);
      ans = ans - tot;
      cout << ans << "\n";
    }
  }
  return 0;
}
