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:
//                I know stuff but probably my rating tells otherwise...

//               Kya hua, code samajhne ki koshish kar rhe ho?? Mat karo,
//                      mujhe bhi samajh nhi aata kya likha hai


import java.io.*;
import java.util.*;

import static java.lang.Math.*;

public class _1291D {

    static void Mangni_ke_bail_ke_dant_na_dekhal_jye() {
        char ch[]=ns().toCharArray();
        n=ch.length;
        int ar[][]=new int[n+5][27];
        for(int i=0;i<n;i++){
            ar[i+1][ch[i]-'a']++;
            for(int j=0;j<27;j++)ar[i+1][j] += ar[i][j];
        }
        t=ni();
        int temp[]=new int[27];
        while(t-->0){
            int l = ni(), r= ni();
            if(l==r){
                pl("Yes");
                continue;
            }
            int count = 0;
            for(int i=0;i<27;i++){
                temp[i] = ar[r][i]- ar[l-1][i];
                if(temp[i]>0)count++;
            }
            if(count ==1)pl("No");
            else if(count >=3)pl("Yes");
            else{
                if(ch[l-1]!=ch[r-1])pl("Yes") ;
                else pl("No");
            }
        }
    }


    //----------------------------------------The main code ends here------------------------------------------------------
    /*-------------------------------------------------------------------------------------------------------------------*/
    //-----------------------------------------Rest's all dust-------------------------------------------------------------


    static int mod9 = 1_000_000_007;
    static int n, m, l, k, t;
    static AwesomeInput input;
    static PrintWriter pw;

    static long power(long a, long b) {
        if (b == 0) return 1;
        if ((b & 1) == 1) return a * power(a * a, b >> 1);
        return power(a * a, b >> 1);
    }

    // The Awesome Input Code is a fast IO method //
    static class AwesomeInput {
        private InputStream letsDoIT;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private SpaceCharFilter filter;

        private AwesomeInput(InputStream incoming) {
            this.letsDoIT = incoming;
        }

        public int read() {
            if (numChars == -1)
                throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = letsDoIT.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        private long ForLong() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            while (!isSpaceChar(c));
            return res * sgn;
        }

        private String ForString() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            }
            while (!isSpaceChar(c));

            return res.toString();
        }

        public boolean isSpaceChar(int c) {
            if (filter != null)
                return filter.isSpaceChar(c);
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public interface SpaceCharFilter {
            boolean isSpaceChar(int ch);
        }
    }

    // functions to take input//
    static int ni() {
        return (int) input.ForLong();
    }

    static String ns() {
        return input.ForString();
    }

    static long nl() {
        return input.ForLong();
    }

    //functions to give output
    static void pl() {
        pw.println();
    }

    static void p(Object o) {
        pw.print(o + " ");
    }

    static void pws(Object o) {
        pw.print(o + "");
    }

    static void pl(Object o) {
        pw.println(o);
    }

    // Fast Sort is Radix Sort
    public static int[] fastSort(int[] f) {
        int n = f.length;
        int[] to = new int[n];
        {
            int[] b = new int[65537];
            for (int i = 0; i < n; i++) b[1 + (f[i] & 0xffff)]++;
            for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
            for (int i = 0; i < n; i++) to[b[f[i] & 0xffff]++] = f[i];
            int[] d = f;
            f = to;
            to = d;
        }
        {
            int[] b = new int[65537];
            for (int i = 0; i < n; i++) b[1 + (f[i] >>> 16)]++;
            for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
            for (int i = 0; i < n; i++) to[b[f[i] >>> 16]++] = f[i];
            int[] d = f;
            f = to;
            to = d;
        }
        return f;
    }

    public static void main(String[] args) {      //threading has been used to increase the stack size.
        try {
            input = new AwesomeInput(System.in);
            pw = new PrintWriter(System.out, true);
            input = new AwesomeInput(new FileInputStream("/home/saurabh/Desktop/input.txt"));
            pw = new PrintWriter(new BufferedWriter(new FileWriter("/home/saurabh/Desktop/output.txt")), true);

        } catch (Exception e) {
        }

        new Thread(null, null, "AApan_gand_hawai_dusar_ke_kare_dawai", 1 << 25)  //the last parameter is stack size desired.
        {
            public void run() {
                try {
                    double s = System.currentTimeMillis();
                    Mangni_ke_bail_ke_dant_na_dekhal_jye();
                    //System.out.println(("\nExecution Time : " + ((double) System.currentTimeMillis() - s) / 1000) + " s");
                    pw.flush();
                    pw.close();
                } catch (Exception e) {
                    e.printStackTrace();
                    System.exit(1);
                }
            }
        }.start();
    }
}