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>
#pragma GCC optimize("Ofast")
using namespace std;
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
struct SegTree_Sum {
  int n;
  vector<long long> Tree, Lazy;
  SegTree_Sum() {}
  SegTree_Sum(int _n) {
    n = _n;
    Tree.resize(n * 4);
    Lazy.resize(n * 4);
  }
  void Propagate(int node, int st, int en) {
    if (Lazy[node] == 0) return;
    Tree[node] += Lazy[node] * (en - st + 1);
    if (st != en) {
      Lazy[node * 2 + 1] += Lazy[node];
      Lazy[node * 2 + 2] += Lazy[node];
    }
    Lazy[node] = 0;
  }
  void Update(int node, int st, int en, int L, int R, long long val) {
    Propagate(node, st, en);
    if (en < st || R < st || en < L) return;
    if (L <= st && en <= R) {
      Lazy[node] += val;
      Propagate(node, st, en);
      return;
    }
    Update(node * 2 + 1, st, (st + en) / 2 + 0, L, R, val);
    Update(node * 2 + 2, (st + en) / 2 + 1, en, L, R, val);
    Tree[node] = Tree[node * 2 + 1] + Tree[node * 2 + 2];
  }
  long long Sum(int node, int st, int en, int L, int R) {
    Propagate(node, st, en);
    if (en < st || R < st || en < L) return 0LL;
    if (L <= st && en <= R) return Tree[node];
    long long p1 = Sum(node * 2 + 1, st, (st + en) / 2 + 0, L, R);
    long long p2 = Sum(node * 2 + 2, (st + en) / 2 + 1, en, L, R);
    return (p1 + p2);
  }
  void RangeUpdate(int L, int R, long long val) {
    Update(0, 0, n - 1, L, R, val);
  }
  long long RangeSum(int L, int R) { return Sum(0, 0, n - 1, L, R); }
};
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n, m, w;
  cin >> n >> m >> w;
  long long TotalCandies = 0;
  SegTree_Sum TreeRow = SegTree_Sum(n);
  SegTree_Sum TreeCol = SegTree_Sum(m);
  while (w--) {
    int t, x1, y1, x2, y2;
    cin >> t >> x1 >> y1 >> x2 >> y2;
    if (t == 0) {
      int v;
      cin >> v;
      TotalCandies += 1LL * (x2 - x1 + 1) * (y2 - y1 + 1) * v;
      TreeRow.RangeUpdate(x1, x2, 1LL * v * (y2 - y1 + 1));
      TreeCol.RangeUpdate(y1, y2, 1LL * v * (x2 - x1 + 1));
    } else if (t == 1) {
      long long res = 0;
      res += TreeRow.RangeSum(x1, x2);
      res += TreeCol.RangeSum(y1, y2);
      res -= TotalCandies;
      cout << res << '\n';
    }
  }
  return 0;
}
