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 javax.print.DocFlavor;
import javax.swing.text.html.HTMLDocument;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.Inet4Address;
import java.nio.charset.IllegalCharsetNameException;
import java.sql.SQLOutput;
import java.util.*;
import java.util.concurrent.locks.ReentrantReadWriteLock;


public class Main {
    static int[] parent;


    public static void main(String[] args) throws IOException{
        Reader.init(System.in);
        int t = 1;
        while (t-->0) {
            solve();
        }




    }


    static  int n ;
    static int m;
    static int k;
    static int[] arr;
    static int[][] dp;
    static void solve()throws IOException {
        String s = Reader.next();
        int q = Reader.nextInt();
        int n = s.length();
        int[][] arr = new int[n+1][26];
        //System.out.println(arr.length);
        for(int i =1 ; i <= n ; i++){
            arr[i][s.charAt(i-1)-'a']++;
            for(int j = 0 ; j < 26 ; j++){
                arr[i][j]+=arr[i-1][j];
            }
        }
        while (q-->0){
            int l = Reader.nextInt();
            int r = Reader.nextInt();
            if (l==r){
                System.out.println("Yes");
                continue;
            }
            int[] tmp = new int[26];
            int cnt = 0;
            int cnt1 = 0;
            int cnt2 = 0;
            for (int i = 0; i < 26; i++) {
                tmp[i] += arr[r][i];
                tmp[i] -= arr[l - 1][i];
                if (tmp[i] > 0) {
                    cnt++;
                }
            }
            if (cnt==2){
                if (s.charAt(l-1)==s.charAt(r-1)){
                    System.out.println("No");
                }
                else{
                    System.out.println("Yes");
                }
            }
            else if (cnt > 1) {
                System.out.println("Yes");
            } else {
                System.out.println("No");
            }

        }

    }








    static int next(long[] arr, long target)
    {
        int start = 0, end = arr.length - 1;

        int ans = -1;
        while (start <= end) {
            int mid = (start + end) / 2;

            // Move to right side if target is
            // greater.
            if (arr[mid] <= target) {
                start = mid + 1;
            }

            // Move left side.
            else {
                ans = mid;
                end = mid - 1;
            }
        }
        return ans;
    }

    static int find(int x) {
        return parent[x] == x ? x : (parent[x] = find(parent[x]));
    }
    static boolean merge(int x, int y) {
        x = find(x);
        y = find(y);
        if (x==y){
            return false;
        }
        parent[x] = y;
        return true;
    }







    public static void sortbyColumn(int arr[][], int col)
    {
        // Using built-in sort function Arrays.sort
        Arrays.sort(arr, new Comparator<int[]>() {

            @Override
            // Compare values according to columns
            public int compare(final int[] entry1,
                               final int[] entry2) {

                // To sort in descending order revert
                // the '>' Operator
                if (entry1[col] > entry2[col])
                    return 1;
                else
                    return -1;
            }
        });  // End of function call sort().
    }


    static class Node{
        long a;
        int i;
        Node(long a, int i){
            this.a = a;
            this.i = i;
        }
    }

}


class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }
    /** get next word */
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                    reader.readLine() );
        }
        return tokenizer.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }

    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
}
class MergeSort
{
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];

        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

    /* A utility function to print array of size n */
    static void printArray(int arr[])
    {
        int n = arr.length;
        for (int i=0; i<n; ++i)
            System.out.print(arr[i] + " ");
        System.out.println();
    }

    // Driver method

}

class Node implements Comparable<Node>{
    int a;
    int b;
    Node (int a , int b){
        this.a = a;
        this.b = b;
    }



    public int compareTo(Node o) {
        if ((this.a%2) == (o.a%2)){
            return (this.b - o.b);
        }
        else{
            return this.a - o.a;
        }
    }
}