#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1010;
vector<int> bl[MAXN][MAXN];
vector<pair<int, int> > rod[MAXN * 2];
long long ans[MAXN * 2];
bool ban[MAXN][MAXN];
void calc(int id) {
  int lst = 0;
  ans[id] = 0;
  for (int i = 0; i < (int)rod[id].size(); i++) {
    int x = rod[id][i].first;
    int y = rod[id][i].second;
    if (ban[x][y])
      lst = i + 1;
    else
      ans[id] += i - lst;
  }
}
int main() {
  int n, m, Q, cnt = 0;
  scanf("%d%d%d", &n, &m, &Q);
  for (int i = 1; i < n; i++) {
    int dx = 1, dy = 0;
    int x = i, y = 1;
    cnt++;
    while (x <= n && y <= m) {
      rod[cnt].push_back(make_pair(x, y));
      bl[x][y].push_back(cnt);
      x += dx, y += dy;
      swap(dx, dy);
    }
  }
  for (int j = 1; j < m; j++) {
    int dx = 0, dy = 1;
    int x = 1, y = j;
    cnt++;
    while (x <= n && y <= m) {
      rod[cnt].push_back(make_pair(x, y));
      bl[x][y].push_back(cnt);
      x += dx, y += dy;
      swap(dx, dy);
    }
  }
  for (int i = 1; i <= cnt; i++) calc(i);
  int x, y;
  long long all = n * m;
  while (Q--) {
    scanf("%d%d", &x, &y);
    if (ban[x][y])
      all++;
    else
      all--;
    ban[x][y] ^= 1;
    for (int id : bl[x][y]) calc(id);
    long long cur = 0;
    for (int i = 1; i <= cnt; i++) cur += ans[i];
    printf("%lld\n", cur + all);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 MAXN = 1010;
vector<int> bl[MAXN][MAXN];
vector<pair<int, int> > rod[MAXN * 2];
long long ans[MAXN * 2];
bool ban[MAXN][MAXN];
void calc(int id) {
  int lst = 0;
  ans[id] = 0;
  for (int i = 0; i < (int)rod[id].size(); i++) {
    int x = rod[id][i].first;
    int y = rod[id][i].second;
    if (ban[x][y])
      lst = i + 1;
    else
      ans[id] += i - lst;
  }
}
int main() {
  int n, m, Q, cnt = 0;
  scanf("%d%d%d", &n, &m, &Q);
  for (int i = 1; i < n; i++) {
    int dx = 1, dy = 0;
    int x = i, y = 1;
    cnt++;
    while (x <= n && y <= m) {
      rod[cnt].push_back(make_pair(x, y));
      bl[x][y].push_back(cnt);
      x += dx, y += dy;
      swap(dx, dy);
    }
  }
  for (int j = 1; j < m; j++) {
    int dx = 0, dy = 1;
    int x = 1, y = j;
    cnt++;
    while (x <= n && y <= m) {
      rod[cnt].push_back(make_pair(x, y));
      bl[x][y].push_back(cnt);
      x += dx, y += dy;
      swap(dx, dy);
    }
  }
  for (int i = 1; i <= cnt; i++) calc(i);
  int x, y;
  long long all = n * m;
  while (Q--) {
    scanf("%d%d", &x, &y);
    if (ban[x][y])
      all++;
    else
      all--;
    ban[x][y] ^= 1;
    for (int id : bl[x][y]) calc(id);
    long long cur = 0;
    for (int i = 1; i <= cnt; i++) cur += ans[i];
    printf("%lld\n", cur + all);
  }
  return 0;
}

```