#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 9;
const long long MOD = 1e9 + 7;
const int mod = 998244353;
inline long long qpow(long long b, long long e, long long m) {
  long long a = 1;
  for (; e; e >>= 1, b = b * b % m)
    if (e & 1) a = a * b % m;
  return a;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, x, y);
  long long z = x;
  x = y, y = z - y * (a / b);
  return d;
}
long long dp[1009][1009][3];
bool a[1009][1009];
int main() {
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] = 1, dp[i][j][1] = 1;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] += dp[i - 1][j][1];
      dp[i][j][1] += dp[i][j - 1][0];
      ans += dp[i][j][0] + dp[i][j][1];
    }
  }
  ans -= n * m;
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[x][y] ^= 1;
    if (a[x][y] == 1) {
      int i = x, j = y, num = 0;
      long long tm = dp[i][j][0];
      while (i <= n && j <= m) {
        if (num % 2 == 0)
          dp[i][j][0] -= tm, j++;
        else
          dp[i][j][1] -= tm, i++;
        ans -= tm;
        num++;
        if (a[i][j] == 1) break;
      }
      i = x, j = y, num = 0;
      tm = dp[i][j][1];
      while (i <= n && j <= m) {
        if (num % 2 == 0)
          dp[i][j][1] -= tm, i++;
        else
          dp[i][j][0] -= tm, j++;
        ans -= tm;
        num++;
        if (a[i][j] == 1) break;
      }
      ans++;
    } else {
      dp[x][y][0] = dp[x - 1][y][1] + 1;
      dp[x][y][1] = dp[x][y - 1][0] + 1;
      int i = x, j = y, num = 1;
      long long tm = dp[x][y][0];
      j++;
      ans += dp[x][y][0];
      while (i <= n && j <= m && a[i][j] == 0) {
        if (num % 2 == 0)
          dp[i][j][0] += tm, j++;
        else
          dp[i][j][1] += tm, i++;
        ans += tm;
        num++;
      }
      i = x, j = y, num = 1;
      tm = dp[x][y][1];
      ans += tm;
      i++;
      while (i <= n && j <= m && a[i][j] == 0) {
        if (num % 2 == 0)
          dp[i][j][1] += tm, i++;
        else
          dp[i][j][0] += tm, j++;
        ans += tm;
        num++;
      }
      ans--;
    }
    printf("%lld\n", ans);
  }
}

### 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 = 1e6 + 9;
const long long MOD = 1e9 + 7;
const int mod = 998244353;
inline long long qpow(long long b, long long e, long long m) {
  long long a = 1;
  for (; e; e >>= 1, b = b * b % m)
    if (e & 1) a = a * b % m;
  return a;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, x, y);
  long long z = x;
  x = y, y = z - y * (a / b);
  return d;
}
long long dp[1009][1009][3];
bool a[1009][1009];
int main() {
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] = 1, dp[i][j][1] = 1;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j][0] += dp[i - 1][j][1];
      dp[i][j][1] += dp[i][j - 1][0];
      ans += dp[i][j][0] + dp[i][j][1];
    }
  }
  ans -= n * m;
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[x][y] ^= 1;
    if (a[x][y] == 1) {
      int i = x, j = y, num = 0;
      long long tm = dp[i][j][0];
      while (i <= n && j <= m) {
        if (num % 2 == 0)
          dp[i][j][0] -= tm, j++;
        else
          dp[i][j][1] -= tm, i++;
        ans -= tm;
        num++;
        if (a[i][j] == 1) break;
      }
      i = x, j = y, num = 0;
      tm = dp[i][j][1];
      while (i <= n && j <= m) {
        if (num % 2 == 0)
          dp[i][j][1] -= tm, i++;
        else
          dp[i][j][0] -= tm, j++;
        ans -= tm;
        num++;
        if (a[i][j] == 1) break;
      }
      ans++;
    } else {
      dp[x][y][0] = dp[x - 1][y][1] + 1;
      dp[x][y][1] = dp[x][y - 1][0] + 1;
      int i = x, j = y, num = 1;
      long long tm = dp[x][y][0];
      j++;
      ans += dp[x][y][0];
      while (i <= n && j <= m && a[i][j] == 0) {
        if (num % 2 == 0)
          dp[i][j][0] += tm, j++;
        else
          dp[i][j][1] += tm, i++;
        ans += tm;
        num++;
      }
      i = x, j = y, num = 1;
      tm = dp[x][y][1];
      ans += tm;
      i++;
      while (i <= n && j <= m && a[i][j] == 0) {
        if (num % 2 == 0)
          dp[i][j][1] += tm, i++;
        else
          dp[i][j][0] += tm, j++;
        ans += tm;
        num++;
      }
      ans--;
    }
    printf("%lld\n", ans);
  }
}

```