#include <bits/stdc++.h>
using namespace std;
int t[2002][2002];
void inc(int d, int i, int delta, int n) {
  for (; i <= n; i |= i + 1) t[d][i] += delta;
}
int sum(int d, int i) {
  int res = 0;
  for (; i > 0; i = (i & (i + 1)) - 1) res += t[d][i];
  return res;
}
bool isFree(int n, int l, int r) { return sum(n, r) == sum(n, l - 1); }
bool a[1001][1001];
int n, m;
int diagonal(int x, int y) { return (n - x) + y; }
int diagonal_index(int x, int y) {
  int k = diagonal(x, y);
  if (k <= n)
    return y;
  else
    return x;
}
int diagonal_size(int x, int y) {
  int k = diagonal(x, y);
  if (k <= n && k <= m) return k;
  if (k <= max(n, m)) return min(n, m);
  return min(n, m) - (k - max(n, m));
}
int get_past(int x, int y) {
  if (x > n || y > m || a[x][y]) return 0;
  int d = diagonal(x, y);
  int i = diagonal_index(x, y);
  int n = diagonal_size(x, y);
  int l = i, r = n, m;
  while (l != r) {
    m = (l + r + 1) / 2;
    if (isFree(d, i, m))
      l = m;
    else
      r = m - 1;
  }
  return l - i + 1;
}
int get_tepa(int x, int y) {
  if (x < 1 || y < 1 || a[x][y]) return 0;
  int d = diagonal(x, y);
  int i = diagonal_index(x, y);
  int l = 1, r = i, m;
  while (l != r) {
    m = (l + r) / 2;
    if (isFree(d, m, i))
      r = m;
    else
      l = m + 1;
  }
  return i - l + 1;
}
void t_main() {
  long long res = 0;
  int q, x, y;
  cin >> n >> m >> q;
  long long Q, W;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < m; j++) {
      res += (diagonal_size(i, j) - diagonal_index(i, j) + 1) +
             (diagonal_size(i + 1, j) - diagonal_index(i + 1, j) + 1);
      res += (diagonal_size(i, j) - diagonal_index(i, j) + 1) +
             (diagonal_size(i, j + 1) - diagonal_index(i, j + 1) + 1);
      res--;
    }
  }
  res += (n + m - 2) * 2 + 1;
  int dia, ind, sz, xyp, xyt, xpyp, xymt, xypp, xmyt;
  bool f;
  for (int i = 0; i < q; i++) {
    cin >> x >> y;
    dia = diagonal(x, y);
    ind = diagonal_index(x, y);
    sz = diagonal_size(x, y);
    if (a[x][y]) {
      f = true;
      inc(dia, ind, -1, sz);
      a[x][y] = 0;
    } else
      f = false;
    xyp = get_past(x, y);
    xyt = get_tepa(x, y);
    xpyp = min(get_past(x + 1, y), xyp);
    xymt = min(get_tepa(x, y - 1), xyt);
    xypp = min(get_past(x, y + 1), xyp);
    xmyt = min(get_tepa(x - 1, y), xyt);
    Q = min(xyp, xpyp + 1) + xpyp;
    Q += (min(xyt, xymt + 1) - 1 + xymt) * Q;
    W = min(xyp, xypp + 1) + xypp;
    W += (min(xyt, xmyt + 1) - 1 + xmyt) * W;
    Q = Q + W - 1;
    if (f) {
      res += Q;
    } else {
      a[x][y] = 1;
      inc(dia, ind, 1, sz);
      res -= Q;
    }
    cout << res << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) t_main();
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int t[2002][2002];
void inc(int d, int i, int delta, int n) {
  for (; i <= n; i |= i + 1) t[d][i] += delta;
}
int sum(int d, int i) {
  int res = 0;
  for (; i > 0; i = (i & (i + 1)) - 1) res += t[d][i];
  return res;
}
bool isFree(int n, int l, int r) { return sum(n, r) == sum(n, l - 1); }
bool a[1001][1001];
int n, m;
int diagonal(int x, int y) { return (n - x) + y; }
int diagonal_index(int x, int y) {
  int k = diagonal(x, y);
  if (k <= n)
    return y;
  else
    return x;
}
int diagonal_size(int x, int y) {
  int k = diagonal(x, y);
  if (k <= n && k <= m) return k;
  if (k <= max(n, m)) return min(n, m);
  return min(n, m) - (k - max(n, m));
}
int get_past(int x, int y) {
  if (x > n || y > m || a[x][y]) return 0;
  int d = diagonal(x, y);
  int i = diagonal_index(x, y);
  int n = diagonal_size(x, y);
  int l = i, r = n, m;
  while (l != r) {
    m = (l + r + 1) / 2;
    if (isFree(d, i, m))
      l = m;
    else
      r = m - 1;
  }
  return l - i + 1;
}
int get_tepa(int x, int y) {
  if (x < 1 || y < 1 || a[x][y]) return 0;
  int d = diagonal(x, y);
  int i = diagonal_index(x, y);
  int l = 1, r = i, m;
  while (l != r) {
    m = (l + r) / 2;
    if (isFree(d, m, i))
      r = m;
    else
      l = m + 1;
  }
  return i - l + 1;
}
void t_main() {
  long long res = 0;
  int q, x, y;
  cin >> n >> m >> q;
  long long Q, W;
  for (int i = 1; i < n; i++) {
    for (int j = 1; j < m; j++) {
      res += (diagonal_size(i, j) - diagonal_index(i, j) + 1) +
             (diagonal_size(i + 1, j) - diagonal_index(i + 1, j) + 1);
      res += (diagonal_size(i, j) - diagonal_index(i, j) + 1) +
             (diagonal_size(i, j + 1) - diagonal_index(i, j + 1) + 1);
      res--;
    }
  }
  res += (n + m - 2) * 2 + 1;
  int dia, ind, sz, xyp, xyt, xpyp, xymt, xypp, xmyt;
  bool f;
  for (int i = 0; i < q; i++) {
    cin >> x >> y;
    dia = diagonal(x, y);
    ind = diagonal_index(x, y);
    sz = diagonal_size(x, y);
    if (a[x][y]) {
      f = true;
      inc(dia, ind, -1, sz);
      a[x][y] = 0;
    } else
      f = false;
    xyp = get_past(x, y);
    xyt = get_tepa(x, y);
    xpyp = min(get_past(x + 1, y), xyp);
    xymt = min(get_tepa(x, y - 1), xyt);
    xypp = min(get_past(x, y + 1), xyp);
    xmyt = min(get_tepa(x - 1, y), xyt);
    Q = min(xyp, xpyp + 1) + xpyp;
    Q += (min(xyt, xymt + 1) - 1 + xymt) * Q;
    W = min(xyp, xypp + 1) + xypp;
    W += (min(xyt, xmyt + 1) - 1 + xmyt) * W;
    Q = Q + W - 1;
    if (f) {
      res += Q;
    } else {
      a[x][y] = 1;
      inc(dia, ind, 1, sz);
      res -= Q;
    }
    cout << res << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) t_main();
  return 0;
}

```