#include <bits/stdc++.h>
using namespace std;
const int M = 1000000007;
const int MM = 998244353;
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
const int N = 1005;
long long dp[N][N][4];
int st[N][N];
int _runtimeTerror_() {
  int n, m, Q;
  cin >> n >> m >> Q;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp[i][j][3] = dp[i][j - 1][2] + 1;
      dp[i][j][2] = dp[i - 1][j][3] + 1;
    }
  }
  for (int i = n; i >= 1; --i) {
    for (int j = m; j >= 1; --j) {
      dp[i][j][0] = dp[i][j + 1][1] + 1;
      dp[i][j][1] = dp[i + 1][j][0] + 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      ans += dp[i][j][0] + dp[i][j][1] - 1;
    }
  }
  auto go = [&](int x, int y) {
    int it = 0;
    int i = x, j = y;
    auto upd1 = [&](int i, int j) {
      if (i < 1 || i > n || j < 1 || j > m) {
        return;
      }
      if (st[i][j] == 1) {
        return;
      }
      dp[i][j][3] = dp[i][j - 1][2] + 1;
      dp[i][j][2] = dp[i - 1][j][3] + 1;
    };
    while (true) {
      upd1(i + 1, j);
      upd1(i, j + 1);
      upd1(i + 1, j + 1);
      i += 1, j += 1;
      if (i > n || j > m) {
        break;
      }
    }
    auto upd2 = [&](int i, int j) {
      if (i < 1 || i > n || j < 1 || j > m) {
        return;
      }
      if (st[i][j] == 1) {
        return;
      }
      dp[i][j][0] = dp[i][j + 1][1] + 1;
      dp[i][j][1] = dp[i + 1][j][0] + 1;
    };
    i = x, j = y;
    while (true) {
      upd2(i - 1, j);
      upd2(i, j - 1);
      upd2(i - 1, j - 1);
      i -= 1, j -= 1;
      if (i < 1 || j < 1) {
        break;
      }
    }
  };
  auto update = [&](int x, int y) {
    if (st[x][y]) {
      int i = x, j = y;
      dp[i][j][3] = dp[i][j - 1][2] + 1;
      dp[i][j][2] = dp[i - 1][j][3] + 1;
      dp[i][j][0] = dp[i][j + 1][1] + 1;
      dp[i][j][1] = dp[i + 1][j][0] + 1;
      ans += dp[i][j][0] * dp[i][j][2];
      ans += dp[i][j][1] * dp[i][j][3];
      ans -= 1;
      go(x, y);
    } else {
      int i = x, j = y;
      ans -= dp[i][j][0] * dp[i][j][2];
      ans -= dp[i][j][1] * dp[i][j][3];
      ans += 1;
      dp[i][j][3] = 0;
      dp[i][j][2] = 0;
      dp[i][j][0] = 0;
      dp[i][j][1] = 0;
      go(x, y);
    }
    st[x][y] ^= 1;
  };
  for (int i = 1; i <= Q; ++i) {
    int x, y;
    cin >> x >> y;
    update(x, y);
    assert(ans >= 0);
    cout << ans << "\n";
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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>
using namespace std;
const int M = 1000000007;
const int MM = 998244353;
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
const int N = 1005;
long long dp[N][N][4];
int st[N][N];
int _runtimeTerror_() {
  int n, m, Q;
  cin >> n >> m >> Q;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp[i][j][3] = dp[i][j - 1][2] + 1;
      dp[i][j][2] = dp[i - 1][j][3] + 1;
    }
  }
  for (int i = n; i >= 1; --i) {
    for (int j = m; j >= 1; --j) {
      dp[i][j][0] = dp[i][j + 1][1] + 1;
      dp[i][j][1] = dp[i + 1][j][0] + 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      ans += dp[i][j][0] + dp[i][j][1] - 1;
    }
  }
  auto go = [&](int x, int y) {
    int it = 0;
    int i = x, j = y;
    auto upd1 = [&](int i, int j) {
      if (i < 1 || i > n || j < 1 || j > m) {
        return;
      }
      if (st[i][j] == 1) {
        return;
      }
      dp[i][j][3] = dp[i][j - 1][2] + 1;
      dp[i][j][2] = dp[i - 1][j][3] + 1;
    };
    while (true) {
      upd1(i + 1, j);
      upd1(i, j + 1);
      upd1(i + 1, j + 1);
      i += 1, j += 1;
      if (i > n || j > m) {
        break;
      }
    }
    auto upd2 = [&](int i, int j) {
      if (i < 1 || i > n || j < 1 || j > m) {
        return;
      }
      if (st[i][j] == 1) {
        return;
      }
      dp[i][j][0] = dp[i][j + 1][1] + 1;
      dp[i][j][1] = dp[i + 1][j][0] + 1;
    };
    i = x, j = y;
    while (true) {
      upd2(i - 1, j);
      upd2(i, j - 1);
      upd2(i - 1, j - 1);
      i -= 1, j -= 1;
      if (i < 1 || j < 1) {
        break;
      }
    }
  };
  auto update = [&](int x, int y) {
    if (st[x][y]) {
      int i = x, j = y;
      dp[i][j][3] = dp[i][j - 1][2] + 1;
      dp[i][j][2] = dp[i - 1][j][3] + 1;
      dp[i][j][0] = dp[i][j + 1][1] + 1;
      dp[i][j][1] = dp[i + 1][j][0] + 1;
      ans += dp[i][j][0] * dp[i][j][2];
      ans += dp[i][j][1] * dp[i][j][3];
      ans -= 1;
      go(x, y);
    } else {
      int i = x, j = y;
      ans -= dp[i][j][0] * dp[i][j][2];
      ans -= dp[i][j][1] * dp[i][j][3];
      ans += 1;
      dp[i][j][3] = 0;
      dp[i][j][2] = 0;
      dp[i][j][0] = 0;
      dp[i][j][1] = 0;
      go(x, y);
    }
    st[x][y] ^= 1;
  };
  for (int i = 1; i <= Q; ++i) {
    int x, y;
    cin >> x >> y;
    update(x, y);
    assert(ans >= 0);
    cout << ans << "\n";
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  return 0;
}

```