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>
inline int inc(int x, int v, int mod) {
  x += v;
  return x >= mod ? x - mod : x;
}
inline int dec(int x, int v, int mod) {
  x -= v;
  return x < 0 ? x + mod : x;
}
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
int _, __, ___;
using namespace std;
template <typename T>
inline void read(T& ret) {
  ret = 0;
  T f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    ret = (ret << 3) + (ret << 1) + (c ^ 48);
    c = getchar();
  }
  ret = ret * f;
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
long long read() {
  long long ret = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    ret = (ret << 3) + (ret << 1) + (c ^ 48);
    c = getchar();
  }
  return ret * f;
}
const int MAXN = 4e6 + 10;
int lowbit(int x) { return x & -x; }
struct bit {
  struct tree {
    int N;
    long long a[MAXN];
    void add(int x, long long v) {
      while (x <= N) a[x] += v, x += lowbit(x);
    }
    long long sum(int x) {
      long long sum = 0;
      while (x) sum += a[x], x -= lowbit(x);
      return sum;
    }
  } T1, T2;
  void add(int x, long long v) {
    T1.add(x, v);
    T2.add(x, x * v);
  }
  void update(int L, int R, long long v) {
    add(L, v);
    add(R + 1, -v);
  }
  long long sum(int x) { return (x + 1) * T1.sum(x) - T2.sum(x); }
  long long query(int L, int R) { return sum(R) - sum(L - 1); }
} tx, ty;
int main() {
  int n, m, w;
  read(n, m, w);
  int op;
  int x1, x2, y1, y2;
  long long v;
  tx.T1.N = tx.T2.N = n;
  ty.T1.N = ty.T2.N = m;
  long long tot = 0;
  while (w--) {
    read(op, x1, y1, x2, y2);
    if (op) {
      long long ans = tx.query(x1, x2) + ty.query(y1, y2);
      printf("%lld\n", ans - tot);
    } else {
      scanf("%lld", &v);
      tot += (x2 - x1 + 1) * (y2 - y1 + 1) * v;
      tx.update(x1, x2, v * (y2 - y1 + 1));
      ty.update(y1, y2, v * (x2 - x1 + 1));
    }
  }
  return 0;
}
