#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e3 + 2;
long long a[maxn][maxn], dp[maxn][maxn], ans[maxn][maxn], tot = 0, n, m;
map<long long, pair<long long, long long> > to;
long long max_st(long long x, long long y) {
  long long d1 = dp[x][y], d2 = dp[x][y + 1];
  d2 = min(d2, d1);
  d1 = min(d1, d2 + 1);
  return d1 + d2;
}
void calc_diag(long long d) {
  long long x = to[d].first, y = to[d].second;
  if (x > n || y > m || x < 1 || y < 1) {
    return;
  }
  while (x >= 1 && y >= 1) {
    if (a[x][y] == 1) {
      dp[x][y] = 0;
    } else {
      dp[x][y] = dp[x + 1][y + 1] + 1;
    }
    x--;
    y--;
  }
}
void calc_ans(long long d) {
  long long x = to[d].first, y = to[d].second;
  if (x > n || y > m || x < 1 || y < 1) {
    return;
  }
  while (x >= 1 && y >= 1) {
    tot -= ans[x][y];
    if (a[x][y]) {
      ans[x][y] = 0;
    } else
      ans[x][y] = max_st(x, y) + max_st(x + 1, y);
    tot += ans[x][y];
    x--;
    y--;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> n >> m >> q;
  for (long long i = 1; i <= n; i++) {
    to[i - m] = {i, m};
    calc_diag(i - m);
  }
  for (long long j = 1; j < m; j++) {
    to[n - j] = {n, j};
    calc_diag(n - j);
  }
  for (long long i = 1; i <= n; i++) {
    calc_ans(i - m);
  }
  for (long long j = 1; j < m; j++) {
    calc_ans(n - j);
  }
  while (q--) {
    long long x, y;
    cin >> x >> y;
    a[x][y] ^= 1;
    calc_diag(x - y);
    calc_diag(x - y - 1);
    calc_diag(x - y + 1);
    calc_ans(x - y);
    calc_ans(x - y - 1);
    calc_ans(x - y + 1);
    cout << tot << "\n";
  }
  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;
const long long maxn = 1e3 + 2;
long long a[maxn][maxn], dp[maxn][maxn], ans[maxn][maxn], tot = 0, n, m;
map<long long, pair<long long, long long> > to;
long long max_st(long long x, long long y) {
  long long d1 = dp[x][y], d2 = dp[x][y + 1];
  d2 = min(d2, d1);
  d1 = min(d1, d2 + 1);
  return d1 + d2;
}
void calc_diag(long long d) {
  long long x = to[d].first, y = to[d].second;
  if (x > n || y > m || x < 1 || y < 1) {
    return;
  }
  while (x >= 1 && y >= 1) {
    if (a[x][y] == 1) {
      dp[x][y] = 0;
    } else {
      dp[x][y] = dp[x + 1][y + 1] + 1;
    }
    x--;
    y--;
  }
}
void calc_ans(long long d) {
  long long x = to[d].first, y = to[d].second;
  if (x > n || y > m || x < 1 || y < 1) {
    return;
  }
  while (x >= 1 && y >= 1) {
    tot -= ans[x][y];
    if (a[x][y]) {
      ans[x][y] = 0;
    } else
      ans[x][y] = max_st(x, y) + max_st(x + 1, y);
    tot += ans[x][y];
    x--;
    y--;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> n >> m >> q;
  for (long long i = 1; i <= n; i++) {
    to[i - m] = {i, m};
    calc_diag(i - m);
  }
  for (long long j = 1; j < m; j++) {
    to[n - j] = {n, j};
    calc_diag(n - j);
  }
  for (long long i = 1; i <= n; i++) {
    calc_ans(i - m);
  }
  for (long long j = 1; j < m; j++) {
    calc_ans(n - j);
  }
  while (q--) {
    long long x, y;
    cin >> x >> y;
    a[x][y] ^= 1;
    calc_diag(x - y);
    calc_diag(x - y - 1);
    calc_diag(x - y + 1);
    calc_ans(x - y);
    calc_ans(x - y - 1);
    calc_ans(x - y + 1);
    cout << tot << "\n";
  }
  return 0;
}

```