Problem: Note that the difference between easy and hard versions is that in hard version unavailable cells can become available again and in easy version can't. You can make hacks only if all versions are solved.

Ildar and Ivan are tired of chess, but they really like the chessboard, so they invented a new game. The field is a chessboard 2n × 2m: it has 2n rows, 2m columns, and the cell in row i and column j is colored white if i+j is even, and is colored black otherwise.

The game proceeds as follows: Ildar marks some of the white cells of the chessboard as unavailable, and asks Ivan to place n × m kings on the remaining white cells in such way, so that there are no kings attacking each other. A king can attack another king if they are located in the adjacent cells, sharing an edge or a corner.

Ildar would like to explore different combinations of cells. Initially all cells are marked as available, and then he has q queries. In each query he marks a cell as unavailable. After each query he would like to know whether it is possible to place the kings on the available cells in a desired way. Please help him!

Input

The first line of input contains three integers n, m, q (1 ≤ n, m, q ≤ 200 000) — the size of the board and the number of queries.

q lines follow, each of them contains a description of a query: two integers i and j, denoting a white cell (i, j) on the board (1 ≤ i ≤ 2n, 1 ≤ j ≤ 2m, i + j is even) that becomes unavailable. It's guaranteed, that each cell (i, j) appears in input at most once.

Output

Output q lines, i-th line should contain answer for a board after i queries of Ildar. This line should contain "YES" if it is possible to place the kings on the available cells in the desired way, or "NO" otherwise.

Examples

Input


1 3 3
1 1
1 5
2 4


Output


YES
YES
NO


Input


3 2 7
4 2
6 4
1 3
2 2
2 4
4 4
3 1


Output


YES
YES
NO
NO
NO
NO
NO

Note

In the first example case after the second query only cells (1, 1) and (1, 5) are unavailable. Then Ivan can place three kings on cells (2, 2), (2, 4) and (2, 6).

After the third query three cells (1, 1), (1, 5) and (2, 4) are unavailable, so there remain only 3 available cells: (2, 2), (1, 3) and (2, 6). Ivan can not put 3 kings on those cells, because kings on cells (2, 2) and (1, 3) attack each other, since these cells share a corner.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int mod = 200003;
const int gmod = 2;
const int inf = 1039074182;
const long long llinf = 1LL * inf * inf;
template <typename T1, typename T2>
inline void chmin(T1 &x, T2 b) {
  if (b < x) x = b;
}
template <typename T1, typename T2>
inline void chmax(T1 &x, T2 b) {
  if (b > x) x = b;
}
template <typename T1, typename T2>
inline void chadd(T1 &x, T2 b) {
  x += b;
  if (x >= mod) x -= mod;
}
template <typename T1, typename T2>
inline void chmul(T1 &x, T2 b) {
  x = 1LL * x * b % mod;
}
template <typename T1, typename T2>
inline void chmod(T1 &x, T2 b) {
  x %= b, x += b;
  if (x >= b) x -= b;
}
using namespace std;
template <typename T>
ostream &operator<<(ostream &cout, vector<T> vec) {
  cout << "{";
  for (int i = 0; i < vec.size(); i++) {
    cout << vec[i];
    if (i != (int)vec.size() - 1) cout << ',';
  }
  cout << "}";
  return cout;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &cout, pair<T1, T2> p) {
  cout << "(" << p.first << ',' << p.second << ")";
  return cout;
}
template <typename T>
ostream &operator<<(ostream &cout, set<T> s) {
  vector<T> t;
  for (auto x : s) t.push_back(x);
  cout << t;
  return cout;
}
template <typename T>
ostream &operator<<(ostream &cout, multiset<T> s) {
  vector<T> t;
  for (auto x : s) t.push_back(x);
  cout << t;
  return cout;
}
template <typename T>
ostream &operator<<(ostream &cout, queue<T> q) {
  vector<T> t;
  while (q.size()) {
    t.push_back(q.front());
    q.pop();
  }
  cout << t;
  return cout;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &cout, map<T1, T2> m) {
  for (auto &x : m) {
    cout << "Key: " << x.first << ' ' << "Value: " << x.second << endl;
  }
  return cout;
}
template <typename T>
T operator*(vector<T> v1, vector<T> v2) {
  assert(v1.size() == v2.size());
  int n = v1.size();
  T res = 0;
  for (int i = 0; i < n; i++) {
    res += v1[i] * v2[i];
  }
  return res;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(pair<T1, T2> x, pair<T1, T2> y) {
  return make_pair(x.first + y.first, x.second + y.second);
}
template <typename T1, typename T2>
pair<T1, T2> operator-(pair<T1, T2> x) {
  return make_pair(-x.first, -x.second);
}
template <typename T>
vector<vector<T> > operator~(vector<vector<T> > vec) {
  vector<vector<T> > v;
  int n = vec.size(), m = vec[0].size();
  v.resize(m);
  for (int i = 0; i < m; i++) {
    v[i].resize(n);
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      v[i][j] = vec[j][i];
    }
  }
  return v;
}
void print0x(int x) {
  std::vector<int> vec;
  while (x) {
    vec.push_back(x & 1);
    x >>= 1;
  }
  std::reverse(vec.begin(), vec.end());
  for (int i = 0; i < vec.size(); i++) {
    std::cout << vec[i];
  }
  std::cout << ' ';
}
template <typename T>
void print0x(T x, int len) {
  std::vector<int> vec;
  while (x) {
    vec.push_back(x & 1);
    x >>= 1;
  }
  reverse(vec.begin(), vec.end());
  for (int i = vec.size(); i < len; i++) {
    putchar('0');
  }
  for (int i = 0; i < vec.size(); i++) {
    std::cout << vec[i];
  }
  std::cout << ' ';
}
using namespace std;
int n, m, q;
set<int> seta[200005];
set<int> setb[200005];
int nn;
int xiaoyu[200005 * 4];
template <typename T>
struct SegmentTreeMinChange {
  const T maxValue = 1e9;
  T *data;
  int nn;
  int size() { return nn; }
  T operator[](int pos) { return data[pos]; }
  void init(int size) {
    nn = 1;
    while (nn < size) {
      nn <<= 1;
    }
    data = new T[nn * 2 + 5];
    for (int i = 0; i <= nn * 2; i++) {
      data[i] = maxValue;
    }
  }
  void update(int pos) { data[pos] = min(data[pos * 2], data[pos * 2 + 1]); }
};
template <typename T>
struct SegmentTreeMaxChange {
  const T minValue = -1e9;
  T *data;
  int nn;
  int size() { return nn; }
  T operator[](int pos) { return data[pos]; }
  void init(int size) {
    nn = 1;
    while (nn < size) {
      nn <<= 1;
    }
    data = new T[nn * 2 + 5];
    for (int i = 0; i <= nn * 2; i++) {
      data[i] = minValue;
    }
  }
  void update(int pos) { data[pos] = max(data[pos * 2], data[pos * 2 + 1]); }
};
SegmentTreeMinChange<int> sgta;
SegmentTreeMaxChange<int> sgtb;
template <typename T>
void change(int pos, int x, T &sgt) {
  pos += nn;
  sgt.data[pos] = x;
  xiaoyu[pos] = sgta[pos] <= sgtb[pos];
  while (pos) {
    pos >>= 1;
    sgt.update(pos);
    xiaoyu[pos] = xiaoyu[pos * 2] | xiaoyu[pos * 2 + 1] |
                  (sgta[pos * 2] <= sgtb[pos * 2 + 1]);
  }
}
int main() {
  cin >> n >> m >> q;
  int x, y;
  sgta.init(n);
  sgtb.init(n);
  memset(xiaoyu, 0, sizeof(xiaoyu));
  nn = 1;
  while (nn < n) nn <<= 1;
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &x, &y);
    x--;
    y--;
    if (x & 1) {
      x /= 2;
      y /= 2;
      if (setb[x].count(y))
        setb[x].erase(y);
      else
        setb[x].insert(y);
      y = (setb[x].empty() ? -inf : *(--setb[x].end()));
      change(x, y, sgtb);
    } else {
      x /= 2;
      y /= 2;
      if (seta[x].count(y))
        seta[x].erase(y);
      else
        seta[x].insert(y);
      y = (seta[x].empty() ? inf : *(seta[x].begin()));
      change(x, y, sgta);
    }
    puts(xiaoyu[1] ? "NO" : "YES");
  }
  return 0;
}