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 ll = long long;
using std::cin;
using std::cout;
using std::endl;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const int inf = (int)1e9 + 7;
const long long INF = 1LL << 60;
template <typename T>
struct BIT {
  int n;
  std::vector<T> bit;
  BIT() : n(0) {}
  BIT(int _n) : n(_n) { bit = std::vector<T>(n + 1); }
  void add1(int idx, T val) {
    for (int i = idx; i <= n; i += i & -i) bit[i] += val;
  }
  T sum1(int idx) {
    T res = 0;
    for (int i = idx; i > 0; i -= i & -i) res += bit[i];
    return res;
  }
  void add(int idx, T val) { add1(idx + 1, val); }
  T sum(int left, int right) { return sum1(right) - sum1(left); }
  int lower_bound(T x) {
    int res = 0;
    int k = 1;
    while (2 * k <= n) k <<= 1;
    for (; k > 0; k >>= 1) {
      if (res + k <= n and bit[res + k] < x) {
        x -= bit[res + k];
        res += k;
      }
    }
    return res;
  }
};
template <typename T>
struct RangeAddBIT {
  int n;
  BIT<T> bit[2];
  RangeAddBIT() : n(0) {}
  RangeAddBIT(int _n) : n(_n) { bit[0] = bit[1] = BIT<T>(n); }
  void add(int left, int right, T val) {
    bit[0].add(left, -val * (left - 1));
    bit[0].add(right, val * (right - 1));
    bit[1].add(left, val);
    bit[1].add(right, -val);
  }
  T sum(int left, int right) {
    T suml = bit[0].sum(0, left) + bit[1].sum(0, left) * (left - 1);
    T sumr = bit[0].sum(0, right) + bit[1].sum(0, right) * (right - 1);
    return sumr - suml;
  }
};
void solve() {
  int n, m, kkt;
  cin >> n >> m >> kkt;
  RangeAddBIT<ll> bh(n), bw(m);
  ll sum = 0;
  while (kkt--) {
    int t;
    cin >> t;
    if (t == 0) {
      int h1, w1, h2, w2, v;
      cin >> h1 >> w1 >> h2 >> w2 >> v;
      h1 -= 1;
      w1 -= 1;
      sum += 1LL * v * (h2 - h1) * (w2 - w1);
      bh.add(h1, h2, v * (w2 - w1));
      bw.add(w1, w2, v * (h2 - h1));
    } else {
      int h1, w1, h2, w2;
      cin >> h1 >> w1 >> h2 >> w2;
      h1 -= 1;
      w1 -= 1;
      cout << bh.sum(h1, h2) + bw.sum(w1, w2) - sum << "\n";
    }
  }
}
int main() {
  std::cin.tie(nullptr);
  std::ios::sync_with_stdio(false);
  int kkt = 1;
  while (kkt--) solve();
  return 0;
}
