#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void chmin(T &a, const T &val) {
  if (a > val) a = val;
}
template <class T>
inline void chmax(T &a, const T &val) {
  if (a < val) a = val;
}
template <class T>
long long sqr(T a) {
  return 1ll * a * a;
}
template <class T>
void compress(T &a) {
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
}
const double pi = acos(-1.00);
const double eps = 1e-6;
const int INF = 2e9;
const int MOD = 1e9 + 7;
const int dx[4] = {0, 0, -1, 1};
const int dy[4] = {-1, 1, 0, 0};
const int N = 1e3 + 1;
struct pack {
  int x, y, k;
};
vector<pack> a[N];
vector<pack> b[N];
vector<int> posA[N][N], posB[N][N];
long long sumA[N], sumB[N];
bool mark[N][N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  for (int j = 1; j <= m; j++) {
    int cur = j;
    for (int i = 1; i <= n; i++) {
      a[j].push_back({i, cur, 0});
      posA[i][cur].push_back(j);
      cur++;
      if (cur > m) break;
      a[j].push_back({i, cur, 0});
      posA[i][cur].push_back(j);
    }
  }
  for (int i = 1; i <= n; i++) {
    int cur = i;
    for (int j = 1; j <= m; j++) {
      b[i].push_back({cur, j, 0});
      posB[cur][j].push_back(i);
      cur++;
      if (cur > n) break;
      b[i].push_back({cur, j, 0});
      posB[cur][j].push_back(i);
    }
  }
  long long tot = n * m;
  for (int i = 1; i <= m; i++) {
    int sz = a[i].size();
    tot += 1ll * sz * (sz - 1) / 2;
    sumA[i] = 1ll * sz * (sz - 1) / 2;
  }
  for (int i = 1; i <= n; i++) {
    int sz = b[i].size();
    tot += 1ll * sz * (sz - 1) / 2;
    sumB[i] = 1ll * sz * (sz - 1) / 2;
  }
  auto updateA = [&](int x, int y, int id) {
    if (id == 0) return;
    for (pack &cur : a[id]) {
      if (cur.x == x && cur.y == y) {
        cur.k = 1 - cur.k;
      }
    }
    tot -= sumA[id];
    int cnt = 0;
    long long tmp = 0;
    for (pack &cur : a[id]) {
      if (cur.k == 1) {
        tmp += 1ll * cnt * (cnt - 1) / 2;
        cnt = 0;
      } else
        cnt++;
    }
    tmp += 1ll * cnt * (cnt - 1) / 2;
    sumA[id] = tmp;
    tot += tmp;
  };
  auto updateB = [&](int x, int y, int id) {
    for (pack &cur : b[id]) {
      if (cur.x == x && cur.y == y) {
        cur.k = 1 - cur.k;
      }
    }
    tot -= sumB[id];
    int cnt = 0;
    long long tmp = 0;
    for (pack &cur : b[id]) {
      if (cur.k == 1) {
        tmp += 1ll * cnt * (cnt - 1) / 2;
        cnt = 0;
      } else
        cnt++;
    }
    tmp += 1ll * cnt * (cnt - 1) / 2;
    sumB[id] = tmp;
    tot += tmp;
  };
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (mark[x][y])
      tot++;
    else
      tot--;
    mark[x][y] = (!mark[x][y]);
    for (int id : posA[x][y]) updateA(x, y, id);
    for (int id : posB[x][y]) updateB(x, y, id);
    cout << tot << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write 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 <class T>
inline void chmin(T &a, const T &val) {
  if (a > val) a = val;
}
template <class T>
inline void chmax(T &a, const T &val) {
  if (a < val) a = val;
}
template <class T>
long long sqr(T a) {
  return 1ll * a * a;
}
template <class T>
void compress(T &a) {
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
}
const double pi = acos(-1.00);
const double eps = 1e-6;
const int INF = 2e9;
const int MOD = 1e9 + 7;
const int dx[4] = {0, 0, -1, 1};
const int dy[4] = {-1, 1, 0, 0};
const int N = 1e3 + 1;
struct pack {
  int x, y, k;
};
vector<pack> a[N];
vector<pack> b[N];
vector<int> posA[N][N], posB[N][N];
long long sumA[N], sumB[N];
bool mark[N][N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  for (int j = 1; j <= m; j++) {
    int cur = j;
    for (int i = 1; i <= n; i++) {
      a[j].push_back({i, cur, 0});
      posA[i][cur].push_back(j);
      cur++;
      if (cur > m) break;
      a[j].push_back({i, cur, 0});
      posA[i][cur].push_back(j);
    }
  }
  for (int i = 1; i <= n; i++) {
    int cur = i;
    for (int j = 1; j <= m; j++) {
      b[i].push_back({cur, j, 0});
      posB[cur][j].push_back(i);
      cur++;
      if (cur > n) break;
      b[i].push_back({cur, j, 0});
      posB[cur][j].push_back(i);
    }
  }
  long long tot = n * m;
  for (int i = 1; i <= m; i++) {
    int sz = a[i].size();
    tot += 1ll * sz * (sz - 1) / 2;
    sumA[i] = 1ll * sz * (sz - 1) / 2;
  }
  for (int i = 1; i <= n; i++) {
    int sz = b[i].size();
    tot += 1ll * sz * (sz - 1) / 2;
    sumB[i] = 1ll * sz * (sz - 1) / 2;
  }
  auto updateA = [&](int x, int y, int id) {
    if (id == 0) return;
    for (pack &cur : a[id]) {
      if (cur.x == x && cur.y == y) {
        cur.k = 1 - cur.k;
      }
    }
    tot -= sumA[id];
    int cnt = 0;
    long long tmp = 0;
    for (pack &cur : a[id]) {
      if (cur.k == 1) {
        tmp += 1ll * cnt * (cnt - 1) / 2;
        cnt = 0;
      } else
        cnt++;
    }
    tmp += 1ll * cnt * (cnt - 1) / 2;
    sumA[id] = tmp;
    tot += tmp;
  };
  auto updateB = [&](int x, int y, int id) {
    for (pack &cur : b[id]) {
      if (cur.x == x && cur.y == y) {
        cur.k = 1 - cur.k;
      }
    }
    tot -= sumB[id];
    int cnt = 0;
    long long tmp = 0;
    for (pack &cur : b[id]) {
      if (cur.k == 1) {
        tmp += 1ll * cnt * (cnt - 1) / 2;
        cnt = 0;
      } else
        cnt++;
    }
    tmp += 1ll * cnt * (cnt - 1) / 2;
    sumB[id] = tmp;
    tot += tmp;
  };
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (mark[x][y])
      tot++;
    else
      tot--;
    mark[x][y] = (!mark[x][y]);
    for (int id : posA[x][y]) updateA(x, y, id);
    for (int id : posB[x][y]) updateB(x, y, id);
    cout << tot << '\n';
  }
  return 0;
}

```