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 either marks a cell as unavailable, or marks the previously unavailable cell as available. 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 on the board (1 ≤ i ≤ 2n, 1 ≤ j ≤ 2m, i + j is even). If the cell (i, j) was available before the query, then it becomes unavailable. Otherwise, if the cell was unavailable, it becomes available.

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 10
4 2
6 4
1 3
4 2
6 4
2 2
2 4
1 3
4 4
3 1


Output


YES
YES
NO
NO
YES
YES
NO
YES
YES
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:
import java.io.*;
import java.util.*;

public class Solution {

    private void solve() throws Exception {
        int n = nextInt();
        int m = nextInt();
        int q = nextInt();

        int nn = n+1;
        int size = 2;
        while (nn > 0) {
            nn /= 2;
            size *= 2;
        }
        a = new int[size];
        Arrays.fill(a, 1<<30);
        b = new int[size];
        Arrays.fill(b, -1);
        flag = new boolean[size];
        Arrays.fill(flag, true);

        TreeSet<Integer>[] A = new TreeSet[n+1];
        TreeSet<Integer>[] B = new TreeSet[n+1];
        for(int i = 0; i <= n; i++) {
            A[i] = new TreeSet<Integer>();
            A[i].add(1<<30);
            B[i] = new TreeSet<Integer>();
            B[i].add(-1);
        }
        
        for (int qq = 1; qq <= q; qq++) {
            int i = nextInt()-1;
            int j = nextInt()-1;
            if(i%2 == 0) {
                int x = i/2;
                int y = j/2;
                if(A[x].contains(y)) A[x].remove(y);
                else A[x].add(y);
                updateIndexA(1, 0, n, x, A[x].first());
            } else {
                int x = i/2+1;
                int y = j/2;
                if(B[x].contains(y)) B[x].remove(y);
                else B[x].add(y);
                updateIndexB(1, 0, n, x, B[x].last());
            }
            System.out.println(flag[1]?"YES":"NO");
//            int jj = 2*m-1-j;
//            if(i%2 == 1) {
//                TreeOdd.updateIndex(1, 0, 2*n-1, i, j);
//                int max = TreeEven.queryRange(1, 0, 2*n-1, 0, i);
//                if(max >= jj) ok = false;
//            } else {
//                TreeEven.updateIndex(1, 0, 2*n-1, i, jj);
//                int max = TreeOdd.queryRange(1, 0, 2*n-1, i, 2*n-1);
//                if(max >= j) ok = false;
//            }
//            System.out.println(ok?"YES":"NO");
        }
    }

    int[] a;
    int[] b;
    boolean[] flag;

    void updateIndexA(int node, int start, int end, int index, int value) {
        if (start > end || start > index || end < index) // Current segment is not within range [l, r]
            return;
        if (start >= index && end <= index) { // Segment is fully within range
            a[node] = value;
            return;
        }
        int mid = (start + end) / 2;
        updateIndexA(node * 2, start, mid, index, value); // Updating left child
        updateIndexA(node * 2 + 1, mid + 1, end, index, value); // Updating right child
        a[node] = Math.min(a[node * 2], a[node * 2 + 1]);
        flag[node] = a[node * 2] > b[node * 2 + 1] && flag[node * 2] && flag[node * 2 + 1];
    }
    
    void updateIndexB(int node, int start, int end, int index, int value) {
        if (start > end || start > index || end < index) // Current segment is not within range [l, r]
            return;
        if (start >= index && end <= index) { // Segment is fully within range
            b[node] = value;
            return;
        }
        int mid = (start + end) / 2;
        updateIndexB(node * 2, start, mid, index, value); // Updating left child
        updateIndexB(node * 2 + 1, mid + 1, end, index, value); // Updating right child
        b[node] = Math.max(b[node * 2], b[node * 2 + 1]);
        flag[node] = a[node * 2] > b[node * 2 + 1] && flag[node * 2] && flag[node * 2 + 1];
    }
    
//    boolean queryRange(int node, int start, int end, int l, int r) {
//        if (start > end || start > r || end < l)
//            return true; // Out of range
//        if (start >= l && end <= r) // Current segment is totally within range [l, r]
//            return flag[node];
//        int mid = (start + end) / 2;
//        boolean V1 = queryRange(node * 2, start, mid, l, r); // Query left child
//        boolean V2 = queryRange(node * 2 + 1, mid + 1, end, l, r); // Query right child
//        return V1 && V2;
//    }
//
//    int queryRangeA(int node, int start, int end, int l, int r) {
//        if (start > end || start > r || end < l)
//            return 1<<30; // Out of range
//        if (start >= l && end <= r) // Current segment is totally within range [l, r]
//            return a[node];
//        int mid = (start + end) / 2;
//        int V1 = queryRangeA(node * 2, start, mid, l, r); // Query left child
//        int V2 = queryRangeA(node * 2 + 1, mid + 1, end, l, r); // Query right child
//        return Math.min(V1, V2);
//    }
//    
//    int queryRangeB(int node, int start, int end, int l, int r) {
//        if (start > end || start > r || end < l)
//            return 0; // Out of range
//        if (start >= l && end <= r) // Current segment is totally within range [l, r]
//            return a[node];
//        int mid = (start + end) / 2;
//        int V1 = queryRangeB(node * 2, start, mid, l, r); // Query left child
//        int V2 = queryRangeB(node * 2 + 1, mid + 1, end, l, r); // Query right child
//        return Math.max(V1, V2);
//    }

    public static void main(String[] args) {
        new Thread(null, new Runnable() {
            public void run() {
                new Solution().run();
            }
        }, "1", 1 << 27).start();
    }

    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer tokenizer;

    public void run() {
        try {
//            in = new BufferedReader(new FileReader("ks_10000_0"));
            in = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
            // out = new PrintWriter(new File("outputPQ.txt"));
            out = new PrintWriter(System.out);
            solve();
            in.close();
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    private int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    private long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    private float nextFloat() throws IOException {
        return Float.parseFloat(nextToken());
    }

    private String nextLine() throws IOException {
        return new String(in.readLine());
    }

    private String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(in.readLine());
        }
        return tokenizer.nextToken();
    }

}