Problem: One day Kefa the parrot was walking down the street as he was on the way home from the restaurant when he saw something glittering by the road. As he came nearer he understood that it was a watch. He decided to take it to the pawnbroker to earn some money. 

The pawnbroker said that each watch contains a serial number represented by a string of digits from 0 to 9, and the more quality checks this number passes, the higher is the value of the watch. The check is defined by three positive integers l, r and d. The watches pass a check if a substring of the serial number from l to r has period d. Sometimes the pawnbroker gets distracted and Kefa changes in some substring of the serial number all digits to c in order to increase profit from the watch. 

The seller has a lot of things to do to begin with and with Kefa messing about, he gave you a task: to write a program that determines the value of the watch.

Let us remind you that number x is called a period of string s (1 ≤ x ≤ |s|), if si = si + x for all i from 1 to |s| - x.

Input

The first line of the input contains three positive integers n, m and k (1 ≤ n ≤ 105, 1 ≤ m + k ≤ 105) — the length of the serial number, the number of change made by Kefa and the number of quality checks.

The second line contains a serial number consisting of n digits.

Then m + k lines follow, containing either checks or changes. 

The changes are given as 1 l r c (1 ≤ l ≤ r ≤ n, 0 ≤ c ≤ 9). That means that Kefa changed all the digits from the l-th to the r-th to be c. 

The checks are given as 2 l r d (1 ≤ l ≤ r ≤ n, 1 ≤ d ≤ r - l + 1).

Output

For each check on a single line print "YES" if the watch passed it, otherwise print "NO".

Examples

Input

3 1 2
112
2 2 3 1
1 1 3 8
2 1 2 1


Output

NO
YES


Input

6 2 3
334934
2 2 5 2
1 4 4 3
2 1 6 3
1 2 3 8
2 3 6 1


Output

NO
YES
NO

Note

In the first sample test two checks will be made. In the first one substring "12" is checked on whether or not it has period 1, so the answer is "NO". In the second one substring "88", is checked on whether or not it has period 1, and it has this period, so the answer is "YES".

In the second statement test three checks will be made. The first check processes substring "3493", which doesn't have period 2. Before the second check the string looks as "334334", so the answer to it is "YES". And finally, the third check processes substring "8334", which does not have period 1.

Solution (in JAVA):

Solution:


import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import static java.lang.Math.*;
import static java.util.Arrays.*;

public class Main {

	public static void main(String[] args) throws IOException {
		new Main().run();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer st = new StringTokenizer("");

	private void run() throws IOException {
		if (new File("input.txt").exists())
			in = new BufferedReader(new FileReader("input.txt"));
		else
			in = new BufferedReader(new InputStreamReader(System.in));
		if (new File("output.txt").exists())
			out = new PrintWriter("output.txt");
		else
			out = new PrintWriter(System.out);
		
		solve();
		
		in.close();
		out.close();
	}
	
	final int P = 73;
	final int MAX_N = 100 * 1000 + 100;
	
	final long POW[] = new long[MAX_N];
	final long POW_S[] = new long[MAX_N];
	
	int n, m;
	Summator s;
	
	int getIndex(char ch) {
	    return 3 + (ch - '0') * 6;
	}
	
	void precalc() {
	    POW[0] = 1;
	    for (int i = 1; i < MAX_N; ++i)
	        POW[i] = POW[i - 1] * P;
	    POW_S[0] = 1;
	    for (int i = 1; i < MAX_N; ++i)
	        POW_S[i] = POW_S[i - 1] + POW[i];
	}
	
	void solve() throws IOException {
	    precalc();
	    
	    n = nextInt();
	    m = nextInt() + nextInt();
	    String text = nextToken();
	    s = new Summator(text);
	    
//	    System.err.println(getIndex('8'));
	    
	    for (int i = 0; i < m; ++i) {
	        int t = nextInt();
	        int l = nextInt() - 1;
	        int r = nextInt() - 1;
	        int v = nextInt();
	        if (t == 2) {
//	            System.err.println(text.substring(l, r + 1));
//	            System.err.println(l + " " + (r - v));
//	            System.err.println(getIndex('1'));
//	            System.err.println(s.getSum(l + v, r));
//	            System.err.println(s.getSum(l + v, r));
	            boolean ans = s.getSum(l + v, r) == s.getSum(l, r - v) * POW[v];
	            if (m == 1 && ans) {
	                for (int d = 0; d < v; ++d)
	                    for (int j = l + d + v; j <= r; j += v)
	                        if (text.charAt(j - v) != text.charAt(j))
	                            ans = false;
	            }
	            out.println(ans ? "YES" : "NO");
	        } else if (t == 1) {
	            s.update(l, r, getIndex((char) ('0' + v)));
	        }
	    }
	}
	
	class Summator {
	    
	    final int n;
	    final long sum[];
	    final int a[];
	    
	    public Summator(String s) {
	        int n = s.length();
	        this.n = Integer.bitCount(n) == 1 ? n : Integer.highestOneBit(n) << 1;
	        sum = new long[this.n << 1];
	        a = new int[this.n << 1];
	        
	        fill(a, 0);
//	        sum[this.n] = getIndex(s.charAt(0));
	        for (int i = 0; i < s.length(); ++i)
	            sum[this.n + i] = getIndex(s.charAt(i)) * POW[i];
	        for (int i = this.n - 1; i > 0; --i)
	            sum[i] = sum[i << 1] + sum[(i << 1) + 1];
        }

	    void push(int v, int l, int r) {
	        if (l > r || a[v] == 0)
	            return;
	        
	        sum[v] = a[v] * (POW_S[r] - (l > 0 ? POW_S[l - 1] : 0));
//	        System.err.println("+ " + l + " " + r + " " + sum[v]);
	        if (l != r)
	            a[v << 1] = a[(v << 1) + 1] = a[v];
	        a[v] = 0;
	    }

	    long getSum(int v, int l, int r, int ll, int rr) {
//	        System.err.println(l + " " + r + " " + ll + " " + rr);
	        push(v, l, r);
	        
	        if (ll > rr)
	            return 0;
	        
	        if (l == ll && r == rr) {
//	            System.err.println(sum[v]);
	            return sum[v];
	        }
	        
	        int m = (l + r) >> 1;
	       	return getSum(v << 1, l, m, ll, min(rr, m)) + getSum((v << 1) + 1, m + 1, r, max(m + 1, ll), rr);        
	    }
	    
	    long getSum(int l, int r) {
	        return getSum(1,  0, n - 1, l, r);
	    }
	    
	    void update(int v, int l, int r, int ll, int rr, int val) {
//	        System.err.println(l + " " + r + " " + ll + " " + rr);
	        push(v, l, r);
	        
	        if (ll > rr)
	            return;
	        
	        if (l == ll && r == rr) {
	            a[v] = val;
	            push(v, l, r);
//	            System.err.println("- " + l + " " + r + " " + sum[v]);
	            return;
	        }
	        
	        int m = (l + r) >> 1;
//	       	System.err.println("m = " + m);
	       	update(v << 1, l, m, ll, min(m, rr), val);
	       	update((v << 1) + 1, m + 1, r, max(m + 1, ll), rr, val);
	       	sum[v] = sum[v << 1] + sum[(v << 1) + 1];
	    }
	    
	    void update(int l, int r, int val) {
	        update(1, 0, n - 1, l, r, val);
	    }
	    
	}

	String nextLine() throws IOException {
		st = new StringTokenizer("");
		return in.readLine();
	}

	String nextToken() throws IOException {
		while (!st.hasMoreTokens()) {
			st = new StringTokenizer(in.readLine());
		}
		return st.nextToken();
	}

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

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

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	boolean EOF() throws IOException {
		while (!st.hasMoreTokens()) {
			String str = in.readLine();
			if (str == null)
				return true;
			st = new StringTokenizer(str);
		}
		return false;
	}

}