#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  long long f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x = x * f;
}
long long n, m, q, s1, s2, ans = 0;
long long sav[120];
long long b[1020][1020];
long long Fquery(long long x, long long y) {
  if (abs(x - y) > 1) return 0;
  if (x == 1 and y == 1) return n * m;
  if (x == y) return 2 * (n - x + 1) * (m - y + 1);
  return (n - x + 1) * (m - y + 1);
}
long long find1(long long s, long long x, long long y, long long dir) {
  if ((dir == 0 and b[x - 1][y]) or (dir == 0 and x == 1)) return s;
  if ((dir == 1 and b[x][y - 1]) or (dir == 1 and y == 1)) return s;
  if (dir == 0)
    return find1(s + 1, x - 1, y, 1);
  else
    return find1(s + 1, x, y - 1, 0);
}
long long find2(long long s, long long x, long long y, long long dir) {
  if ((dir == 0 and b[x + 1][y]) or (dir == 0 and x == n)) return s;
  if ((dir == 1 and b[x][y + 1]) or (dir == 1 and y == m)) return s;
  if (dir == 0)
    return find2(s + 1, x + 1, y, 1);
  else
    return find2(s + 1, x, y + 1, 0);
}
signed main() {
  read(n), read(m), read(q);
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) ans += Fquery(i, j);
  for (long long i = 1; i <= q; i++) {
    read(s1), read(s2);
    sav[1] = find1(1, s1, s2, 0);
    sav[2] = find1(1, s1, s2, 1);
    sav[3] = find2(1, s1, s2, 0);
    sav[4] = find2(1, s1, s2, 1);
    if (b[s1][s2] == 0)
      b[s1][s2] = 1, ans -= sav[1] * sav[4] + sav[2] * sav[3], ans++;
    else
      b[s1][s2] = 0, ans += sav[1] * sav[4] + sav[2] * sav[3], ans--;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution 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;
template <typename T>
void read(T &x) {
  x = 0;
  long long f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x = x * f;
}
long long n, m, q, s1, s2, ans = 0;
long long sav[120];
long long b[1020][1020];
long long Fquery(long long x, long long y) {
  if (abs(x - y) > 1) return 0;
  if (x == 1 and y == 1) return n * m;
  if (x == y) return 2 * (n - x + 1) * (m - y + 1);
  return (n - x + 1) * (m - y + 1);
}
long long find1(long long s, long long x, long long y, long long dir) {
  if ((dir == 0 and b[x - 1][y]) or (dir == 0 and x == 1)) return s;
  if ((dir == 1 and b[x][y - 1]) or (dir == 1 and y == 1)) return s;
  if (dir == 0)
    return find1(s + 1, x - 1, y, 1);
  else
    return find1(s + 1, x, y - 1, 0);
}
long long find2(long long s, long long x, long long y, long long dir) {
  if ((dir == 0 and b[x + 1][y]) or (dir == 0 and x == n)) return s;
  if ((dir == 1 and b[x][y + 1]) or (dir == 1 and y == m)) return s;
  if (dir == 0)
    return find2(s + 1, x + 1, y, 1);
  else
    return find2(s + 1, x, y + 1, 0);
}
signed main() {
  read(n), read(m), read(q);
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) ans += Fquery(i, j);
  for (long long i = 1; i <= q; i++) {
    read(s1), read(s2);
    sav[1] = find1(1, s1, s2, 0);
    sav[2] = find1(1, s1, s2, 1);
    sav[3] = find2(1, s1, s2, 0);
    sav[4] = find2(1, s1, s2, 1);
    if (b[s1][s2] == 0)
      b[s1][s2] = 1, ans -= sav[1] * sav[4] + sav[2] * sav[3], ans++;
    else
      b[s1][s2] = 0, ans += sav[1] * sav[4] + sav[2] * sav[3], ans--;
    cout << ans << endl;
  }
  return 0;
}

```