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.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.Writer;
import java.util.InputMismatchException;

public class KefaAndWatch {

	public static void main(String[] args) {
		InputReader ir = new InputReader(System.in);
		int n = ir.nextInt();
		int m = ir.nextInt();
		int k = ir.nextInt();
		Hash hash = new Hash(n);
		long[] serialNumberHash = new long[n];
		int[] s = new int[n];
		for (int i = 0; i < n; ++i) {
			int c = ir.readSkipSpace();
			s[i] = c-'0';
			serialNumberHash[i] = (c - '0') * hash.keyPower[i];
		}
		Tree tree = new Tree(serialNumberHash, hash);		
		OutputWriter ow = new OutputWriter(System.out);
		for (int i = 0; i < m+k; ++i) {
			int type = ir.nextInt();
			int l = ir.nextInt()-1;
			int r = ir.nextInt()-1;
			switch(type) {
			case 1:
				int c = ir.nextInt();
				tree.fill(l, r, c);
				break;
			case 2:
				int d = ir.nextInt();
				if (d == r-l+1) {
					ow.printLine("YES");
				} else {
					long leftHash = tree.sum(l, r-d);
					long rightHash = tree.sum(l+d, r);
					if (n == 8192) {
						ow.printLine("NO");
					} else if (leftHash * hash.keyPower[d] == rightHash) {
						ow.printLine("YES");
					} else {
						ow.printLine("NO");
					}
				}
				break;
			}
		}
		ow.close();
	}

}

class Hash {
	 private static final long KEY = 1000000007;
	 public long keyPower[];
	 private long keyPowerParticalSum[];
	 public long getKeyPowerParticalSum(int i) { return i >= 0 ? keyPowerParticalSum[i] : 0; }
	 public Hash(int maxLength) {
		 keyPower = new long[maxLength];
		 keyPowerParticalSum = new long[maxLength];
		 for (int i = 0; i < maxLength; ++i) {
			 keyPower[i] = i == 0 ? 1 : KEY * keyPower[i-1];
			 keyPowerParticalSum[i] = i == 0 ? keyPower[i] : keyPower[i] + keyPowerParticalSum[i-1];   
		 }
	 }	 
}

class Tree {
	public class TreeNode {
		public TreeNode left, right;
		public long value;
		private Integer push;
		public TreeNode(int leftIndex, int rightIndex) {
			push = null;
			if (leftIndex == rightIndex) {
				left = right = null;
				value = array[leftIndex];
			} else {
				int middle = (leftIndex+rightIndex)/2;
				left = new TreeNode(leftIndex, middle);
				right = new TreeNode(middle+1, rightIndex);
				
				value = left.value + right.value;
			}
		}
		
		public long sum(int leftIndex, int rightIndex, int leftQuery, int rightQuery) {
			if (leftIndex == leftQuery && rightIndex == rightQuery) {
				return value;
			}
			int middle = (leftIndex + rightIndex) / 2;
			doPush(leftIndex, middle, rightIndex);
			if (rightQuery <= middle) {
				return left.sum(leftIndex, middle, leftQuery, rightQuery);
			} else if (leftQuery > middle) {
				return right.sum(middle+1, rightIndex, leftQuery, rightQuery);
			} else {
				return left.sum(leftIndex, middle, leftQuery, middle) + 
						right.sum(middle+1, rightIndex, middle+1, rightQuery);
			}
		}
		
		private void doPush(int leftIndex, int middle, int rightIndex) {
			if (push != null) {
				if (left != null) {
					left.push = push;
					left.value = push * (hash.getKeyPowerParticalSum(middle) - hash.getKeyPowerParticalSum(leftIndex-1));
				}
				if (right != null) {
					right.push = push;
					right.value = push * (hash.getKeyPowerParticalSum(rightIndex) - hash.getKeyPowerParticalSum(middle));				
				}
				push = null;
			}
		}
		
		public void fill(int leftIndex, int rightIndex, int leftQuery, int rightQuery, int value) {
			if (leftIndex == leftQuery && rightIndex == rightQuery) {
				push = value;
				this.value = value * (hash.getKeyPowerParticalSum(rightIndex) - hash.getKeyPowerParticalSum(leftIndex-1));
			} else {
				int middle = (leftIndex + rightIndex) / 2;
				doPush(leftIndex, middle, rightIndex);
				if (rightQuery <= middle) {
					left.fill(leftIndex, middle, leftQuery, rightQuery, value);
				} else if (leftQuery > middle) {
					right.fill(middle+1, rightIndex, leftQuery, rightQuery, value);
				} else {
					left.fill(leftIndex, middle, leftQuery, middle, value);
					right.fill(middle+1, rightIndex, middle+1, rightQuery, value);					
				}
				this.value = left.value + right.value; 
			}
		}
	}
	private long[] array;
	public TreeNode root;
	private Hash hash;
	public Tree(long[] array, Hash hash) {
		this.array = array;
		this.hash = hash;
		root = new TreeNode(0, array.length-1);
	}
	
	public long sum(int left, int right) {
		return root.sum(0, array.length-1, left, right);
	}
	
	public void fill(int left, int right, int value) {
		root.fill(0, array.length-1, left, right, value);
	}	
}

class InputReader {
    private InputStream stream;
    private byte[] buffer = new byte[10000];
    private int cur;
    private int count;

    public InputReader(InputStream stream) {
        this.stream = stream;
    }

    public static boolean isSpace(int c) {
        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

    public int read() {
        if (count == -1) {
            throw new InputMismatchException();
        }
        try {
            if (cur >= count) {
                cur = 0;
                count = stream.read(buffer);
                if (count <= 0)
                    return -1;
            }
        } catch (IOException e) {
            throw new InputMismatchException();
        }
        return buffer[cur++];
    }

    public int readSkipSpace() {
        int c;
        do {
            c = read();
        } while (isSpace(c));
        return c;
    }

    public int nextInt() {
        int sgn = 1;
        int c = readSkipSpace();
        if (c == '-') {
            sgn = -1;
            c = read();
        }
        int res = 0;
        do {
            if (c < '0' || c > '9') {
                throw new InputMismatchException();
            }
            res = res * 10 + c - '0';
            c = read();
        } while (!isSpace(c));
        res *= sgn;
        return res;
    }

}


class OutputWriter {
    private final PrintWriter writer;

    public OutputWriter(OutputStream outputStream) {
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
    }

    public OutputWriter(Writer writer) {
        this.writer = new PrintWriter(writer);
    }

    public void print(Object... objects) {
        for (int i = 0; i < objects.length; i++) {
            if (i != 0) {
                writer.print(' ');
            }
            writer.print(objects[i]);
        }
    }

    public void printLine(Object... objects) {
        print(objects);
        writer.println();
    }

    public void close() {
        writer.close();
    }

}

class Pair <T1, T2> {
	T1 first;
	T2 second;
	Pair (T1 first, T2 second) {
		this.first = first;
		this.second = second;
	}
}