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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long int myRand(long long int B) { return (unsigned long long)rng() % B; }
struct LazySegmentTree {
 private:
  int n;
  vector<long long int> node, lazy;

 public:
  LazySegmentTree(vector<long long int> v) {
    int sz = (int)v.size();
    n = 1;
    while (n < sz) n *= 2;
    node.resize(2 * n - 1);
    lazy.resize(2 * n - 1, 0);
    for (int i = 0; i < sz; i++) node[i + n - 1] = v[i];
    for (int i = n - 2; i >= 0; i--)
      node[i] = node[i * 2 + 1] + node[i * 2 + 2];
  }
  void eval(int k, int l, int r) {
    if (lazy[k] != 0) {
      node[k] += lazy[k];
      if (r - l > 1) {
        lazy[2 * k + 1] += lazy[k] / 2;
        lazy[2 * k + 2] += lazy[k] / 2;
      }
      lazy[k] = 0;
    }
  }
  void add(int a, int b, long long int x, int k = 0, int l = 0, int r = -1) {
    if (r < 0) r = n;
    eval(k, l, r);
    if (b <= l || r <= a) return;
    if (a <= l && r <= b) {
      lazy[k] += (r - l) * x;
      eval(k, l, r);
    } else {
      add(a, b, x, 2 * k + 1, l, (l + r) / 2);
      add(a, b, x, 2 * k + 2, (l + r) / 2, r);
      node[k] = node[2 * k + 1] + node[2 * k + 2];
    }
  }
  long long int getsum(int a, int b, int k = 0, int l = 0, int r = -1) {
    if (r < 0) r = n;
    eval(k, l, r);
    if (b <= l || r <= a) return 0;
    if (a <= l && r <= b) return node[k];
    long long int vl = getsum(a, b, 2 * k + 1, l, (l + r) / 2);
    long long int vr = getsum(a, b, 2 * k + 2, (l + r) / 2, r);
    return vl + vr;
  }
};
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n, m, q;
  cin >> n >> m >> q;
  LazySegmentTree seg1(vector<long long int>(n, 0));
  LazySegmentTree seg2(vector<long long int>(m, 0));
  long long int sum = 0;
  while (q--) {
    int t;
    cin >> t;
    int x, y, xx, yy;
    cin >> x >> y >> xx >> yy;
    if (t == 0) {
      long long int v;
      cin >> v;
      sum += v * (long long int)(xx - x + 1) * (long long int)(yy - y + 1);
      seg1.add(x - 1, xx, v * (yy - y + 1));
      seg2.add(y - 1, yy, v * (xx - x + 1));
    } else {
      long long int res = -sum;
      res += seg1.getsum(x - 1, xx);
      res += seg2.getsum(y - 1, yy);
      printf("%lld\n", res);
    }
  }
}
