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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
/*
 */
public class B {

	static long[] xCoords, yCoords;
	
	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		char[] string=fs.next().toCharArray();
		int nQueries=fs.nextInt();
		int n=string.length;
		int[][] cs=new int[n+1][26];
		for (int i=1; i<=n; i++) {
			for (int x=0; x<26; x++) {
				cs[i][x]=cs[i-1][x]+(string[i-1]-'a' == x?1:0);
			}
		}
		
		PrintWriter out=new PrintWriter(System.out);
		for (int qq=0; qq<nQueries; qq++) {
			int l=fs.nextInt()-1, r=fs.nextInt()-1;
			if (l==r) {
				out.println("Yes");
			}
			else if (string[l]!=string[r]) {
				out.println("Yes");
			}
			else {
				//otherwise we need at least three different letters
				int nonzeroCount=0;
				for (int x=0; x<26; x++) {
					if (cs[r+1][x]-cs[l][x]>0) nonzeroCount++;
				}
				if (nonzeroCount>=3) {
					out.println("Yes");
				}
				else {
					out.println("No");
				}
			}
		}
		out.close();
	}
	
	static class HarmeyerHash implements Comparable<HarmeyerHash> {
		static final long m1=8675309, m2=1_000_000_007;
		long v1=0, v2=0; int l=0;	
		static final long s1=257, s2=619;
		static long[] s1Pow, s2Pow;
		static boolean precomped=false;
		
		void add(char o) {
			v1=(v1*s1+o)%m1;
			v2=(v2*s2+o)%m2;
			l++;
		}
		
		public int compareTo(HarmeyerHash o) {
			if (v1!=o.v1)
				return Long.compare(v1, o.v1);
				return Long.compare(v2, o.v2);
		}
		
		public boolean equals(Object o) {
			return compareTo((HarmeyerHash)o)==0;
		}

		public int hashCode() {
			return (int)v1;
		}
		
		public HarmeyerHash clone() {
			HarmeyerHash toReturn=new HarmeyerHash();
			toReturn.v1=v1;
			toReturn.v2=v2;
			toReturn.l=l;
			return toReturn;
		}

		static void precomp() {
			if (precomped) return;
			precomped=true;
			s1Pow=new long[1000_000];
			s2Pow=new long[1000_000];
			s1Pow[0]=s2Pow[0]=1;
			for (int i=1; i<s1Pow.length; i++)
				s1Pow[i]=(s1Pow[i-1]*s1)%m1;
			for (int i=1; i<s2Pow.length; i++)
				s2Pow[i]=(s2Pow[i-1]*s2)%m2;
		}

		//need fastPow if o can be longer than 10^6
		void append(HarmeyerHash o) {
			precomp();
			v1=(v1*s1Pow[o.l]+o.v1)%m1;
			v2=(v2*s2Pow[o.l]+o.v2)%m2;
			l+=o.l;
		}
		
		public static HarmeyerHash[] getPrefixHashes(char[] word) {
			precomp();
			int n=word.length;
			HarmeyerHash[] toReturn=new HarmeyerHash[n+1];
			toReturn[0]=new HarmeyerHash();
			for (int i=1; i<=n; i++) {
				toReturn[i]=toReturn[i-1].clone();
				toReturn[i].add(word[i-1]);
			}
			return toReturn;
		}
		
		//inclusive, exclusive
		public static HarmeyerHash substringHash(HarmeyerHash[] prefixHashes, int from, int to) {
			if (from==to)
				return new HarmeyerHash();
			HarmeyerHash old=prefixHashes[to].clone(), toSub=prefixHashes[from];
			int diff=to-from;
			old.v1=(old.v1-(toSub.v1*s1Pow[diff])%m1+m1)%m1;
			old.v2=(old.v2-(toSub.v2*s2Pow[diff])%m2+m2)%m2;
			old.l=to-from;
			return old;
		}
		
	}

	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		
		String next() {
			while (!st.hasMoreElements())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}

}