#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T, class second>
ostream &operator<<(ostream &os, const pair<T, second> &v) {
  os << v.first << ' ' << v.second;
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (T i : v) os << i << ' ';
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &v) {
  for (T i : v) os << i << ' ';
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const map<T, second> &v) {
  for (auto i : v) os << '(' << i.first << "," << i.second << ')' << ' ';
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const unordered_map<T, second> &v) {
  for (auto i : v) os << '(' << i.first << "," << i.second << ')' << ' ';
  return os;
}
long long m1(vector<vector<int> > &r, vector<vector<int> > &d,
             vector<vector<int> > &s, int x, int y, long long c, int f,
             int fl) {
  long long t = c;
  bool f2 = true;
  while (x < r.size() && y < r[0].size() && f2) {
    if (f == 1) {
      f--;
      x++;
      if (x < r.size() && s[x][y] == 0) {
        d[x][y] += fl * c;
        t += c;
      }
      if (x < r.size() && s[x][y] == 1) {
        f2 = false;
      }
    } else {
      f++;
      y++;
      if (y < r[0].size() && s[x][y] == 0) {
        t += c;
        r[x][y] += fl * c;
      }
      if (y < r[0].size() && s[x][y] == 1) {
        f2 = false;
      }
    }
  }
  return t;
}
void solve() {
  int n, m, q;
  cin >> n >> m >> q;
  long long a = -1 * n * m;
  vector<vector<int> > s(n, vector<int>(m, 0));
  vector<vector<int> > r(n, vector<int>(m, 1)), d(n, vector<int>(m, 1));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (j > 0) {
        r[i][j] += d[i][j - 1];
      }
      if (i > 0) {
        d[i][j] += r[i - 1][j];
      }
      a += r[i][j] + d[i][j];
    }
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    if (s[x][y] == 0) {
      s[x][y] = 1;
      long long t = m1(r, d, s, x, y, r[x][y], 1, -1);
      t += m1(r, d, s, x, y, d[x][y], 0, -1);
      t--;
      a -= t;
      r[x][y] = 0;
      d[x][y] = 0;
    } else {
      s[x][y] = 0;
      r[x][y] = 1;
      d[x][y] = 1;
      if (x > 0) {
        d[x][y] += r[x - 1][y];
      }
      if (y > 0) {
        r[x][y] += d[x][y - 1];
      }
      long long t = m1(r, d, s, x, y, r[x][y], 1, 1);
      t += m1(r, d, s, x, y, d[x][y], 0, 1);
      t--;
      a += t;
    }
    cout << a;
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T, class second>
ostream &operator<<(ostream &os, const pair<T, second> &v) {
  os << v.first << ' ' << v.second;
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (T i : v) os << i << ' ';
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &v) {
  for (T i : v) os << i << ' ';
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const map<T, second> &v) {
  for (auto i : v) os << '(' << i.first << "," << i.second << ')' << ' ';
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const unordered_map<T, second> &v) {
  for (auto i : v) os << '(' << i.first << "," << i.second << ')' << ' ';
  return os;
}
long long m1(vector<vector<int> > &r, vector<vector<int> > &d,
             vector<vector<int> > &s, int x, int y, long long c, int f,
             int fl) {
  long long t = c;
  bool f2 = true;
  while (x < r.size() && y < r[0].size() && f2) {
    if (f == 1) {
      f--;
      x++;
      if (x < r.size() && s[x][y] == 0) {
        d[x][y] += fl * c;
        t += c;
      }
      if (x < r.size() && s[x][y] == 1) {
        f2 = false;
      }
    } else {
      f++;
      y++;
      if (y < r[0].size() && s[x][y] == 0) {
        t += c;
        r[x][y] += fl * c;
      }
      if (y < r[0].size() && s[x][y] == 1) {
        f2 = false;
      }
    }
  }
  return t;
}
void solve() {
  int n, m, q;
  cin >> n >> m >> q;
  long long a = -1 * n * m;
  vector<vector<int> > s(n, vector<int>(m, 0));
  vector<vector<int> > r(n, vector<int>(m, 1)), d(n, vector<int>(m, 1));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (j > 0) {
        r[i][j] += d[i][j - 1];
      }
      if (i > 0) {
        d[i][j] += r[i - 1][j];
      }
      a += r[i][j] + d[i][j];
    }
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    if (s[x][y] == 0) {
      s[x][y] = 1;
      long long t = m1(r, d, s, x, y, r[x][y], 1, -1);
      t += m1(r, d, s, x, y, d[x][y], 0, -1);
      t--;
      a -= t;
      r[x][y] = 0;
      d[x][y] = 0;
    } else {
      s[x][y] = 0;
      r[x][y] = 1;
      d[x][y] = 1;
      if (x > 0) {
        d[x][y] += r[x - 1][y];
      }
      if (y > 0) {
        r[x][y] += d[x][y - 1];
      }
      long long t = m1(r, d, s, x, y, r[x][y], 1, 1);
      t += m1(r, d, s, x, y, d[x][y], 0, 1);
      t--;
      a += t;
    }
    cout << a;
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}

```