#include <bits/stdc++.h>
using namespace std;
long long n, m, qq, w, h, sum, p, q;
bool vis[4105][4105];
pair<long long, long long> te;
set<long long> s[4105];
long long get(long long h, long long w, long long tag) {
  long long sum = 0;
  if (tag == 1) {
    sum += 2050;
  }
  if (h != 1)
    sum += 1020 + h;
  else if (w != 1)
    sum += w;
  return sum;
}
pair<long long, long long> hd(long long h, long long w, long long tag) {
  long long d = min(h - 1, w - 1);
  h -= d;
  w -= d;
  long long cnt = 2 * d;
  if (tag == 1 && h > 1) {
    cnt++;
    h--;
    tag = 2;
  } else if (tag == 2 && w > 1) {
    cnt++;
    w--;
    tag = 1;
  }
  return make_pair(get(h, w, tag), cnt + 1);
}
void ini(long long h, long long w, long long tag) {
  te = hd(h, w, tag);
  s[te.first].insert(te.second + 1);
  s[te.first].insert(0);
  sum += (te.second + 1) * te.second / 2;
}
void add(long long tag) {
  te = hd(h, w, tag);
  s[te.first].erase(te.second);
  p = *(s[te.first].upper_bound(te.second));
  q = *(prev(s[te.first].lower_bound(te.second)));
  sum += (te.second - q) * (p - te.second);
}
void del(long long tag) {
  te = hd(h, w, tag);
  p = *(s[te.first].upper_bound(te.second));
  q = *(prev(s[te.first].lower_bound(te.second)));
  sum -= (te.second - q) * (p - te.second);
  s[te.first].insert(te.second);
}
signed main() {
  scanf("%lld%lld%lld", &n, &m, &qq);
  for (long long i = 1; i <= n; i++) {
    ini(i, m, 1);
  }
  for (long long j = 1; j <= m; j++) {
    ini(n, j, 2);
  }
  sum -= n * m;
  while (qq--) {
    scanf("%lld%lld", &h, &w);
    if (vis[h][w]) {
      add(1);
      add(2);
      sum--;
      vis[h][w] = 0;
    } else {
      del(1);
      del(2);
      sum++;
      vis[h][w] = 1;
    }
    printf("%lld\n", sum);
  }
  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>
using namespace std;
long long n, m, qq, w, h, sum, p, q;
bool vis[4105][4105];
pair<long long, long long> te;
set<long long> s[4105];
long long get(long long h, long long w, long long tag) {
  long long sum = 0;
  if (tag == 1) {
    sum += 2050;
  }
  if (h != 1)
    sum += 1020 + h;
  else if (w != 1)
    sum += w;
  return sum;
}
pair<long long, long long> hd(long long h, long long w, long long tag) {
  long long d = min(h - 1, w - 1);
  h -= d;
  w -= d;
  long long cnt = 2 * d;
  if (tag == 1 && h > 1) {
    cnt++;
    h--;
    tag = 2;
  } else if (tag == 2 && w > 1) {
    cnt++;
    w--;
    tag = 1;
  }
  return make_pair(get(h, w, tag), cnt + 1);
}
void ini(long long h, long long w, long long tag) {
  te = hd(h, w, tag);
  s[te.first].insert(te.second + 1);
  s[te.first].insert(0);
  sum += (te.second + 1) * te.second / 2;
}
void add(long long tag) {
  te = hd(h, w, tag);
  s[te.first].erase(te.second);
  p = *(s[te.first].upper_bound(te.second));
  q = *(prev(s[te.first].lower_bound(te.second)));
  sum += (te.second - q) * (p - te.second);
}
void del(long long tag) {
  te = hd(h, w, tag);
  p = *(s[te.first].upper_bound(te.second));
  q = *(prev(s[te.first].lower_bound(te.second)));
  sum -= (te.second - q) * (p - te.second);
  s[te.first].insert(te.second);
}
signed main() {
  scanf("%lld%lld%lld", &n, &m, &qq);
  for (long long i = 1; i <= n; i++) {
    ini(i, m, 1);
  }
  for (long long j = 1; j <= m; j++) {
    ini(n, j, 2);
  }
  sum -= n * m;
  while (qq--) {
    scanf("%lld%lld", &h, &w);
    if (vis[h][w]) {
      add(1);
      add(2);
      sum--;
      vis[h][w] = 0;
    } else {
      del(1);
      del(2);
      sum++;
      vis[h][w] = 1;
    }
    printf("%lld\n", sum);
  }
  return 0;
}

```