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":

<image>

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.util.*;
import java.io.*;
public class Main{


	public static void main (String args[])throws IOException{
		Scanner sc = new Scanner();
		PrintWriter out = new PrintWriter(System.out);
		char[] str = sc.next().toCharArray(); 
		int t =sc.nextInt(), arr[][]=new int[str.length][26], k;
		arr[0][str[0]-97]=1;
		for(int i=1; i<str.length; i++) {
			for(int j=0; j<26; j++)
				arr[i][j] = arr[i-1][j];
			arr[i][str[i]-97]++;
		}
		
		while(t-->0) {
			int start=sc.nextInt()-1, end=sc.nextInt()-1, arr1[]=new int[26];
			if(start==end || str[start]!=str[end])
				out.println("YES");
			else {
				k=0;
				if(start!=0) {
					for(int j=0; j<26 && k<3; j++)
						if(arr[end][j]-arr[start-1][j]>0)
							k++;
				}
				else {
					for(int j=0; j<26 && k<3; j++)
						if(arr[end][j]>0)
							k++;
				}
				
					
				
				if(k==3)
					out.println("YES");
				else
					out.println("NO");
			}
			
		
			out.flush();
		}
		
	}	

	
	
	
	
	
	
	
	
	
	

	static long power(long num, long power) {
		long temp=1;
		for(int i=0; i<power; i++) {
			temp*=num;
			if(temp>100000)
				return -1;
		}
		return temp;
	}
	
	static long nearestKthPower(long num, long k) {
		long temp=1, count=0, a=num;
		
		while(num%2==0) {
			num/=2;
			count++;
		}
		if(count>=k)
			count=count%k;
		else if(count!=0)
			count=k-count;
		while(count-->0) {
			temp*=2;
			if(temp>100000)
				return -1;
		}
		if(num==1)
			return temp;
		for(int i=3; i*i<=a; i+=2) {
			count=0;
			while(num%i==0) {
				num/=i;
				count++;
			}
			if(count>=k)
				count=count%k;
			else if(count!=0)
				count=k-count;
			while(count-->0) {
				temp*=i;
				if(temp>100000)
					return -1;
			}
			if(num==1)
				return temp;
		}
		if(num>1) {
			count=k-1;
			while(count-->0) {
				temp*=num;
				if(temp>100000)
					return -1;
			}
		}
		return temp;
			
	}
	
    static int[] primeGenerator(int num) {
    	int length=0, arr[]=new int[num], a=num, factor=1;
    	if(num%2==0) {
    		while(num%2==0) {
    			num/=2;
    			factor*=2;
    		}
    		arr[length++]=factor;
    	}
    	for(int i=3; i*i<=a; i++) {
    		factor=1;
    		if(num%i==0) {
        		while(num%i==0) {
        			num/=i;
        			factor*=i;
        		}
        		arr[length++]=factor;
    		}
    	}
    	if(num>1)
    		arr[length++]=num;
    	return Arrays.copyOfRange(arr, 0, length);
    }


    static boolean isPrime(int n) 
    { 
        // Corner cases 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
  
        // This is checked so that we can skip 
        // middle five numbers in below loop 
        if (n % 2 == 0 || n % 3 == 0) 
            return false; 
  
        for (int i = 5; i * i <= n; i = i + 6) 
            if (n % i == 0 || n % (i + 2) == 0) 
                return false; 
  
        return true; 
    }












	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		Scanner(String fileName) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(fileName));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		String nextLine() throws IOException {
			return br.readLine();
		}

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

		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

		boolean ready() throws IOException {
			return br.ready();
		}

	}

	static void sort(int[] a) {
		shuffle(a);
		Arrays.sort(a);
	}

	static void shuffle(int[] a) {
		int n = a.length;
		Random rand = new Random();
		for (int i = 0; i < n; i++) {
			int tmpIdx = rand.nextInt(n);
			int tmp = a[i];
			a[i] = a[tmpIdx];
			a[tmpIdx] = tmp;
		}
	}
}