You are given a directed graph consisting of n vertices. Each directed edge (or arc) labeled with a single character. Initially, the graph is empty.

You should process m queries with it. Each query is one of three types: 

  * "+ u v c" — add arc from u to v with label c. It's guaranteed that there is no arc (u, v) in the graph at this moment; 
  * "- u v" — erase arc from u to v. It's guaranteed that the graph contains arc (u, v) at this moment; 
  * "? k" — find the sequence of k vertices v_1, v_2, ..., v_k such that there exist both routes v_1 → v_2 → ... → v_k and v_k → v_{k - 1} → ... → v_1 and if you write down characters along both routes you'll get the same string. You can visit the same vertices any number of times. 

Input

The first line contains two integers n and m (2 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ m ≤ 2 ⋅ 10^5) — the number of vertices in the graph and the number of queries.

The next m lines contain queries — one per line. Each query is one of three types: 

  * "+ u v c" (1 ≤ u, v ≤ n; u ≠ v; c is a lowercase Latin letter); 
  * "- u v" (1 ≤ u, v ≤ n; u ≠ v); 
  * "? k" (2 ≤ k ≤ 10^5). 



It's guaranteed that you don't add multiple edges and erase only existing edges. Also, there is at least one query of the third type.

Output

For each query of the third type, print YES if there exist the sequence v_1, v_2, ..., v_k described above, or NO otherwise.

Example

Input


3 11
+ 1 2 a
+ 2 3 b
+ 3 2 a
+ 2 1 b
? 3
? 2
- 2 1
- 3 2
+ 2 1 c
+ 3 2 d
? 5


Output


YES
NO
YES

Note

In the first query of the third type k = 3, we can, for example, choose a sequence [1, 2, 3], since 1 \xrightarrow{a} 2 \xrightarrow{b} 3 and 3 \xrightarrow{a} 2 \xrightarrow{b} 1.

In the second query of the third type k = 2, and we can't find sequence p_1, p_2 such that arcs (p_1, p_2) and (p_2, p_1) have the same characters.

In the third query of the third type, we can, for example, choose a sequence [1, 2, 3, 2, 1], where 1 \xrightarrow{a} 2 \xrightarrow{b} 3 \xrightarrow{d} 2 \xrightarrow{c} 1.
import java.io.*;
import java.util.*;

public class Main{
	static void main() throws Exception{
		int n=sc.nextInt(),m=sc.nextInt();
		TreeSet<int[]>pairs=new TreeSet<int[]>((x,y)->x[0]==y[0]?x[1]-y[1]:x[0]-y[0]);
		TreeSet<int[]>eqPairs=new TreeSet<int[]>((x,y)->x[0]==y[0]?x[1]-y[1]:x[0]-y[0]);
		TreeMap<int[],Character>ts=new TreeMap<int[],Character>((x,y)->x[0]==y[0]?x[1]-y[1]:x[0]-y[0]);
		while(m-->0) {
			char cmd=sc.nextChar();
			if(cmd=='?') {
				int k=sc.nextInt();
				if(pairs.isEmpty()) {
					pw.println("NO");
					continue;
				}
				if(((k&1)==1) || !eqPairs.isEmpty()) {
					pw.println("YES");
					continue;
				}
				pw.println("NO");
				continue;
			}
			int u=sc.nextInt(),v=sc.nextInt();
			int x=Math.min(u, v),y=Math.max(u, v);
			if(cmd=='-') {
				pairs.remove(new int[] {x,y});
				eqPairs.remove(new int[] {x,y});
				ts.remove(new int[] {u,v});
				continue;
			}
			char c=sc.nextChar();
			if(ts.containsKey(new int[] {v,u})) {
				pairs.add(new int[] {x,y});
				if(ts.get(new int[] {v,u})==c)
					eqPairs.add(new int[] {x,y});
			}
			ts.put(new int[] {u,v}, c);
		}
	}
	public static void main(String[] args) throws Exception{
		sc=new MScanner(System.in);
		pw = new PrintWriter(System.out);
		int tc=1;
//		tc=sc.nextInt();
		for(int i=1;i<=tc;i++) {
//			pw.printf("Case %d:\n", i);
			main();
		}
		pw.flush();
	}
	static PrintWriter pw;
	static MScanner sc;
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
	 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
	 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] intArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] longArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public int[] intSortedArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public long[] longSortedArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public Integer[] IntegerArr(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] LongArr(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
	 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
	 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
	 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
	 
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
	 
		public boolean ready() throws IOException {
			return br.ready();
		}
	 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
		
	}
	static void shuffle(int[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			int tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
	static void shuffle(long[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			long tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
}