Problem: Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.

Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:

  1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s; 
  2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t; 
  3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other. 



If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.

For example, consider the string s =  "gamegame". Then the string t =  "megamage" is a reducible anagram of s, we may choose for example s_1 =  "game", s_2 =  "gam", s_3 =  "e" and t_1 =  "mega", t_2 =  "mag", t_3 =  "e":

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On the other hand, we can prove that t =  "memegaga" is an irreducible anagram of s.

You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.

Input

The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).

The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.

Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.

Output

For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.

Examples

Input


aaaaa
3
1 1
2 4
5 5


Output


Yes
No
Yes


Input


aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5


Output


No
Yes
Yes
Yes
No
No

Note

In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 =  "a", s_2 =  "aa", t_1 =  "a", t_2 =  "aa" to show that it is a reducible anagram.

In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.InputStreamReader;
import java.io.BufferedOutputStream;
import java.util.StringTokenizer;
import java.io.Closeable;
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.Flushable;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	static class TaskAdapter implements Runnable {
		@Override
		public void run() {
			InputStream inputStream = System.in;
			OutputStream outputStream = System.out;
			Input in = new Input(inputStream);
			Output out = new Output(outputStream);
			BIrreducibleAnagrams solver = new BIrreducibleAnagrams();
			solver.solve(1, in, out);
			out.close();
		}
	}

	public static void main(String[] args) throws Exception {
		Thread thread = new Thread(null, new TaskAdapter(), "", 1<<29);
		thread.start();
		thread.join();
	}
	static class BIrreducibleAnagrams {
		public BIrreducibleAnagrams() {
		}

		public void solve(int kase, Input in, Output pw) {
			char[] input = in.next().toCharArray();
			int n = input.length;
			int[][] psum = new int[n+1][26];
			for(int i = 1; i<=n; i++) {
				System.arraycopy(psum[i-1], 0, psum[i], 0, 26);
				psum[i][input[i-1]-'a']++;
			}
			int q = in.nextInt();
			while(q-->0) {
				int l = in.nextInt()-1, r = in.nextInt()-1;
				if(r-l+1<=1||input[l]!=input[r]) {
					pw.println("Yes");
					continue;
				}
				int cnt = 0;
				for(int i = 0; i<26; i++) {
					if(psum[r+1][i]-psum[l][i]>0) {
						cnt++;
					}
				}
				if(cnt>=3) {
					pw.println("Yes");
				}else {
					pw.println("No");
				}
			}
		}

	}

	static class Input {
		BufferedReader br;
		StringTokenizer st;

		public Input(InputStream is) {
			this(is, 1<<20);
		}

		public Input(InputStream is, int bs) {
			br = new BufferedReader(new InputStreamReader(is), bs);
			st = null;
		}

		public boolean hasNext() {
			try {
				while(st==null||!st.hasMoreTokens()) {
					String s = br.readLine();
					if(s==null) {
						return false;
					}
					st = new StringTokenizer(s);
				}
				return true;
			}catch(Exception e) {
				return false;
			}
		}

		public String next() {
			if(!hasNext()) {
				throw new InputMismatchException();
			}
			return st.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

	}

	static class Output implements Closeable, Flushable {
		public StringBuilder sb;
		public OutputStream os;
		public int BUFFER_SIZE;
		public boolean autoFlush;
		public String LineSeparator;

		public Output(OutputStream os) {
			this(os, 1<<16);
		}

		public Output(OutputStream os, int bs) {
			BUFFER_SIZE = bs;
			sb = new StringBuilder(BUFFER_SIZE);
			this.os = new BufferedOutputStream(os, 1<<17);
			autoFlush = false;
			LineSeparator = System.lineSeparator();
		}

		public void println(String s) {
			sb.append(s);
			println();
			if(autoFlush) {
				flush();
			}else if(sb.length()>BUFFER_SIZE >> 1) {
				flushToBuffer();
			}
		}

		public void println() {
			sb.append(LineSeparator);
		}

		private void flushToBuffer() {
			try {
				os.write(sb.toString().getBytes());
			}catch(IOException e) {
				e.printStackTrace();
			}
			sb = new StringBuilder(BUFFER_SIZE);
		}

		public void flush() {
			try {
				flushToBuffer();
				os.flush();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

		public void close() {
			flush();
			try {
				os.close();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

	}
}