#include <bits/stdc++.h>
using namespace std;
template <class T>
void chmax(T& a, const T& b) {
  a = max(a, b);
}
template <class T>
void chmin(T& a, const T& b) {
  a = min(a, b);
}
using ll = long long;
using P = pair<int, int>;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VVL = vector<VL>;
const int dx[2] = {1, 0};
const int dy[2] = {0, 1};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  while (tt--) {
    int n, m, q;
    cin >> n >> m >> q;
    vector<set<int>> d;
    vector<vector<vector<pair<int, int>>>> data_ij(
        n, vector<vector<pair<int, int>>>(m));
    vector<vector<char>> state(n, vector<char>(m));
    ll now = 0;
    auto count = [](ll x, ll y) {
      x++;
      if (y - x < 2) return 0LL;
      if (x & 1) x++;
      ll n = y - x;
      if (n % 2 == 0) {
        n /= 2;
        return n * n;
      } else
        return (n + 1) / 2 * ((n - 1) / 2);
    };
    for (int k0 = 0; k0 < int(2); ++k0) {
      for (int i = n - 1, j = 0; j < m;) {
        int idx = d.size();
        d.resize(idx + 1);
        auto& st = d[idx];
        st.insert(-1);
        int pos = 0;
        for (int x = i, y = j, k = k0; x < n && y < m;
             x += dx[k], y += dy[k], k ^= 1) {
          data_ij[x][y].emplace_back(idx, pos++);
        }
        st.insert(pos);
        now += count(-1, pos);
        if (i)
          i--;
        else
          j++;
      }
    }
    now += n * m;
    while (q--) {
      int i, j;
      cin >> i >> j;
      i--, j--;
      if (state[i][j]) {
        now++;
        for (auto [idx, pos] : data_ij[i][j]) {
          auto& st = d[idx];
          auto it = st.lower_bound(pos);
          auto pit = prev(it);
          auto nit = next(it);
          now -= count(*pit, *it);
          now -= count(*it, *nit);
          now += count(*pit, *nit);
          st.erase(it);
        }
      } else {
        now--;
        for (auto [idx, pos] : data_ij[i][j]) {
          auto& st = d[idx];
          auto nit = st.lower_bound(pos);
          auto pit = prev(nit);
          now -= count(*pit, *nit);
          now += count(*pit, pos);
          now += count(pos, *nit);
          st.insert(pos);
        }
      }
      state[i][j] ^= 1;
      cout << now << '\n';
    }
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

You are given a matrix, consisting of n rows and m columns. The rows are numbered top to bottom, the columns are numbered left to right.

Each cell of the matrix can be either free or locked.

Let's call a path in the matrix a staircase if it: 

  * starts and ends in the free cell; 
  * visits only free cells; 
  * has one of the two following structures: 
    1. the second cell is 1 to the right from the first one, the third cell is 1 to the bottom from the second one, the fourth cell is 1 to the right from the third one, and so on; 
    2. the second cell is 1 to the bottom from the first one, the third cell is 1 to the right from the second one, the fourth cell is 1 to the bottom from the third one, and so on. 



In particular, a path, consisting of a single cell, is considered to be a staircase.

Here are some examples of staircases:

<image>

Initially all the cells of the matrix are free.

You have to process q queries, each of them flips the state of a single cell. So, if a cell is currently free, it makes it locked, and if a cell is currently locked, it makes it free.

Print the number of different staircases after each query. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.

Input

The first line contains three integers n, m and q (1 ≤ n, m ≤ 1000; 1 ≤ q ≤ 10^4) — the sizes of the matrix and the number of queries.

Each of the next q lines contains two integers x and y (1 ≤ x ≤ n; 1 ≤ y ≤ m) — the description of each query.

Output

Print q integers — the i-th value should be equal to the number of different staircases after i queries. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.

Examples

Input


2 2 8
1 1
1 1
1 1
2 2
1 1
1 2
2 1
1 1


Output


5
10
5
2
5
3
1
0


Input


3 4 10
1 4
1 2
2 3
1 2
2 3
3 2
1 3
3 4
1 3
3 1


Output


49
35
24
29
49
39
31
23
29
27


Input


1000 1000 2
239 634
239 634


Output


1332632508
1333333000

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
void chmax(T& a, const T& b) {
  a = max(a, b);
}
template <class T>
void chmin(T& a, const T& b) {
  a = min(a, b);
}
using ll = long long;
using P = pair<int, int>;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VVL = vector<VL>;
const int dx[2] = {1, 0};
const int dy[2] = {0, 1};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  while (tt--) {
    int n, m, q;
    cin >> n >> m >> q;
    vector<set<int>> d;
    vector<vector<vector<pair<int, int>>>> data_ij(
        n, vector<vector<pair<int, int>>>(m));
    vector<vector<char>> state(n, vector<char>(m));
    ll now = 0;
    auto count = [](ll x, ll y) {
      x++;
      if (y - x < 2) return 0LL;
      if (x & 1) x++;
      ll n = y - x;
      if (n % 2 == 0) {
        n /= 2;
        return n * n;
      } else
        return (n + 1) / 2 * ((n - 1) / 2);
    };
    for (int k0 = 0; k0 < int(2); ++k0) {
      for (int i = n - 1, j = 0; j < m;) {
        int idx = d.size();
        d.resize(idx + 1);
        auto& st = d[idx];
        st.insert(-1);
        int pos = 0;
        for (int x = i, y = j, k = k0; x < n && y < m;
             x += dx[k], y += dy[k], k ^= 1) {
          data_ij[x][y].emplace_back(idx, pos++);
        }
        st.insert(pos);
        now += count(-1, pos);
        if (i)
          i--;
        else
          j++;
      }
    }
    now += n * m;
    while (q--) {
      int i, j;
      cin >> i >> j;
      i--, j--;
      if (state[i][j]) {
        now++;
        for (auto [idx, pos] : data_ij[i][j]) {
          auto& st = d[idx];
          auto it = st.lower_bound(pos);
          auto pit = prev(it);
          auto nit = next(it);
          now -= count(*pit, *it);
          now -= count(*it, *nit);
          now += count(*pit, *nit);
          st.erase(it);
        }
      } else {
        now--;
        for (auto [idx, pos] : data_ij[i][j]) {
          auto& st = d[idx];
          auto nit = st.lower_bound(pos);
          auto pit = prev(nit);
          now -= count(*pit, *nit);
          now += count(*pit, pos);
          now += count(pos, *nit);
          st.insert(pos);
        }
      }
      state[i][j] ^= 1;
      cout << now << '\n';
    }
  }
}

```