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;
struct FenwickTree {
  void Init(int _n) {
    n = _n;
    a.assign(n, 0);
  }
  long long Get(int pos) const {
    long long r = 0;
    for (; pos >= 0; pos = (pos & (pos + 1)) - 1) {
      r += a[pos];
    }
    return r;
  }
  void Inc(int pos, long long val) {
    for (; pos < n; pos |= (pos + 1)) {
      a[pos] += val;
    }
  }
  long long Get(int l, int r) const {
    if (l == 0) {
      return Get(r);
    }
    return Get(r) - Get(l - 1);
  }
  int n;
  vector<long long> a;
};
struct RangeFenwickTree {
  void Init(int _n) {
    n = _n;
    add.Init(_n);
    mult.Init(_n);
  }
  long long Get(int pos) const {
    long long ca = add.Get(pos);
    long long cm = mult.Get(pos);
    return cm * pos + ca;
  }
  long long Get(int l, int r) const {
    if (l == 0) {
      return Get(r);
    }
    return Get(r) - Get(l - 1);
  }
  void Inc(int l, int r, long long val) {
    add.Inc(l, -val * (l - 1));
    mult.Inc(l, val);
    if (r + 1 < n) {
      mult.Inc(r + 1, -val);
      add.Inc(r + 1, val * r);
    }
  }
  int n;
  FenwickTree add, mult;
};
void TestRangeFenwickTree() {
  int n;
  cin >> n;
  vector<long long> a(n, 0);
  RangeFenwickTree fw;
  fw.Init(n);
  for (int itr = 0; itr < 100; ++itr) {
    int l = rand() % n;
    int r = rand() % n;
    if (l > r) {
      swap(l, r);
    }
    long long val = rand();
    fw.Inc(l, r, val);
    for (int i = l; i <= r; ++i) {
      a[i] += val;
    }
    for (int i = 0; i < n; ++i) {
      long long sum = 0;
      for (int j = i; j < n; ++j) {
        sum += a[j];
        assert(fw.Get(i, j) == sum);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m, w;
  cin >> n >> m >> w;
  RangeFenwickTree sx, sy;
  sx.Init(n);
  sy.Init(m);
  while (w > 0) {
    int tp;
    int x1, y1, x2, y2;
    cin >> tp >> x1 >> y1 >> x2 >> y2;
    --x1;
    --y1;
    --x2;
    --y2;
    if (tp == 0) {
      long long val;
      cin >> val;
      sx.Inc(x1, x2, val * (y2 - y1 + 1));
      sy.Inc(y1, y2, val * (x2 - x1 + 1));
    } else {
      long long res = sx.Get(x1, x2) + sy.Get(y1, y2);
      res -= sx.Get(0, x1 - 1);
      res -= sx.Get(x2 + 1, n - 1);
      res -= sy.Get(0, y1 - 1);
      res -= sy.Get(y2 + 1, m - 1);
      cout << res / 2 << '\n';
    }
    --w;
  }
  cout.flush();
  return 0;
}
