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 JAVA):

Solution:
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
What do you think? What do you think?
1st on Billboard, what do you think of it
Next is a Grammy, what do you think of it
However you think, I’m sorry, but shit, I have no fcking interest
*******************************
I'm standing on top of my Monopoly board
That means I'm on top of my game and it don't stop
til my hip don't hop anymore
https://www.a2oj.com/Ladder16.html
*******************************
300iq as writer = Sad!
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class F1
   {
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         int M = Integer.parseInt(st.nextToken());
         int Q = Integer.parseInt(st.nextToken());
         Point[] arr = new Point[Q];
         for(int i=0; i < Q; i++)
         {
            st = new StringTokenizer(infile.readLine());
            int x = Integer.parseInt(st.nextToken());
            int y = Integer.parseInt(st.nextToken());
            arr[i] = new Point(x,y);
         }
         int low = 0;
         int high = Q-1;
         while(low != high)
         {
            int mid = (low+high+1)/2;
            int[] top = new int[2*N+1];
            Arrays.fill(top, 4*M);
            SegmentTree bottom = new SegmentTree(1, 2*N);
            for(int i=0; i <= mid; i++)
            {
               int x = arr[i].x;
               int y = arr[i].y;
               if(x%2 == 0 && bottom.query(x-1,x-1) < y-1)
                  bottom.update(x-1, y-1);
               else if(x%2 == 1)
                  top[x] = Math.min(top[x], y);
            }
            boolean works = true;
            for(int a=1; a <= 2*N; a+=2)
               if(top[a] <= bottom.query(a, 2*N))
                  works = false;
            if(works)
               low = mid;
            else
               high = mid-1;
         }
         StringBuilder sb = new StringBuilder();
         for(int i=0; i <= low; i++)
            sb.append("YES\n");
         for(int i=low+1; i < Q; i++)
            sb.append("NO\n");
         System.out.print(sb);
      }
   }
   class Point
   {
      public int x;
      public int y;
      
      public Point(int a, int b)
      {
         x = a;
         y = b;
      }
   }
   class SegmentTree 
   {
       //range query, non lazy
       final int[] val;
       final int treeFrom;
       final int length;
   
       public SegmentTree(int treeFrom, int treeTo) 
       {
           this.treeFrom = treeFrom;
           int length = treeTo - treeFrom + 1;
           int l;
           for (l = 0; (1 << l) < length; l++);
           val = new int[1 << (l + 1)];
           this.length = 1 << l;
       }
       public void update(int index, int delta) 
       {
          //replaces value
          int node = index - treeFrom + length;
          val[node] = delta;
          for (node >>= 1; node > 0; node >>= 1) 
            val[node] = comb(val[node << 1], val[(node << 1) + 1]);
       }
       public int query(int from, int to) 
       {
          //inclusive bounds
          if (to < from)
             return 0; //0 or 1?
          from += length - treeFrom;
          to += length - treeFrom + 1;
          //0 or 1?
          int res = 0;
          for (; from + (from & -from) <= to; from += from & -from) 
              res = comb(res, val[from / (from & -from)]);
          for (; to - (to & -to) >= from; to -= to & -to)
              res = comb(res, val[(to - (to & -to)) / (to & -to)]);
          return res;
       }
       public int comb(int a, int b)
       {
          //change this
          return Math.max(a,b);
       }
   }