#include <bits/stdc++.h>
using namespace std;
const long long N = 1e3 + 10;
long long n, m, Q, res, cnt;
long long a[N][N], f[N][N];
inline void query(long long x, long long y, long long tag) {
  if (x ^ n) {
    long long cur = min(f[x + 1][y], f[x][y]);
    res += tag * cur;
    if (f[x + 1][y] >= f[x][y])
      res += max(0LL, f[x][y] - 1) * tag;
    else
      res += tag * f[x + 1][y];
  }
  if (y ^ m) {
    long long cur = min(f[x][y + 1], f[x][y]);
    res += tag * cur;
    if (f[x][y + 1] >= f[x][y])
      res += max(0LL, f[x][y] - 1) * tag;
    else
      res += tag * f[x][y + 1];
  }
  return;
}
inline void modify(long long x, long long y) {
  if (a[x][y])
    --cnt;
  else
    ++cnt;
  a[x][y] ^= 1;
  if (a[x][y])
    f[x][y] = f[x + 1][y + 1] + 1;
  else
    f[x][y] = 0;
  return;
}
inline void upd(long long x, long long y, long long tag) {
  long long l = x, r = y;
  while (l && r) query(l, r, tag), --l, --r;
  return;
}
inline void update(long long x, long long y) {
  upd(x, y, -1), upd(x - 1, y, -1);
  upd(x, y - 1, -1);
  modify(x, y);
  long long l = x - 1, r = y - 1;
  while (l && r) {
    if (a[l][r])
      f[l][r] = f[l + 1][r + 1] + 1;
    else
      f[l][r] = 0;
    --l, --r;
  }
  upd(x, y, 1);
  upd(x - 1, y, 1);
  upd(x, y - 1, 1);
  return;
}
signed main() {
  scanf("%lld%lld%lld", &n, &m, &Q);
  cnt = n * m;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j) a[i][j] = 1;
  for (long long i = n; i; --i)
    for (long long j = m; j; --j) f[i][j] = f[i + 1][j + 1] + 1;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j) query(i, j, 1);
  while (Q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    update(x, y);
    printf("%lld\n", res + cnt);
  }
  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 N = 1e3 + 10;
long long n, m, Q, res, cnt;
long long a[N][N], f[N][N];
inline void query(long long x, long long y, long long tag) {
  if (x ^ n) {
    long long cur = min(f[x + 1][y], f[x][y]);
    res += tag * cur;
    if (f[x + 1][y] >= f[x][y])
      res += max(0LL, f[x][y] - 1) * tag;
    else
      res += tag * f[x + 1][y];
  }
  if (y ^ m) {
    long long cur = min(f[x][y + 1], f[x][y]);
    res += tag * cur;
    if (f[x][y + 1] >= f[x][y])
      res += max(0LL, f[x][y] - 1) * tag;
    else
      res += tag * f[x][y + 1];
  }
  return;
}
inline void modify(long long x, long long y) {
  if (a[x][y])
    --cnt;
  else
    ++cnt;
  a[x][y] ^= 1;
  if (a[x][y])
    f[x][y] = f[x + 1][y + 1] + 1;
  else
    f[x][y] = 0;
  return;
}
inline void upd(long long x, long long y, long long tag) {
  long long l = x, r = y;
  while (l && r) query(l, r, tag), --l, --r;
  return;
}
inline void update(long long x, long long y) {
  upd(x, y, -1), upd(x - 1, y, -1);
  upd(x, y - 1, -1);
  modify(x, y);
  long long l = x - 1, r = y - 1;
  while (l && r) {
    if (a[l][r])
      f[l][r] = f[l + 1][r + 1] + 1;
    else
      f[l][r] = 0;
    --l, --r;
  }
  upd(x, y, 1);
  upd(x - 1, y, 1);
  upd(x, y - 1, 1);
  return;
}
signed main() {
  scanf("%lld%lld%lld", &n, &m, &Q);
  cnt = n * m;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j) a[i][j] = 1;
  for (long long i = n; i; --i)
    for (long long j = m; j; --j) f[i][j] = f[i + 1][j + 1] + 1;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j) query(i, j, 1);
  while (Q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    update(x, y);
    printf("%lld\n", res + cnt);
  }
  return 0;
}

```