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:
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

import java.io.BufferedReader;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 *
 * @author is2ac
 */
public class D_CF {
    public static boolean fail;
    public static int id;
    public static void main(String[] args) throws IOException {
        FastScanner58 fs = new FastScanner58();
        //Reader fs = new Reader();
        PrintWriter pw = new PrintWriter(System.out);
        //int t = fs.ni();
        int t = 1;
        for (int tc = 0; tc < t; tc++) {
            String s = fs.next();
            int n = s.length();
            int q = fs.ni();
            int[][] a = new int[n][26];
            for (int i = 0; i < n; i++) {
                a[i][s.charAt(i)-'a']++;
            }
            for(int i = 1; i < n; i++) {
                for (int j = 0; j < 26; j++) {
                    a[i][j] += a[i-1][j];
                }
            }
            for (int i = 0; i < q; i++) {
                int l = fs.ni()-1, r = fs.ni()-1;
                int c = 0;
                int let = 0, max = (int)(-1e9);
                boolean flag = true;
                for (int j = 0; j < 26; j++) {
                    let = a[r][j];
                    if (l>0) let -= a[l-1][j];
                    if (let>0) c++;
                    max = Math.max(max,let);
                }
                if (c==2) {
                    if (s.charAt(l)==s.charAt(r)) {
                        pw.println("No");
                    } else {
                        pw.println("Yes");
                    }
                } else if (c>2 || l==r) {
                    pw.println("Yes");
                } else {
                    pw.println("No");
                }
            }
            for (int i = 0; i < n; i++) {
                //pw.println(Arrays.toString(a[i]));
            }
        }
        pw.close();
    }
    public static long gcd(long n1, long n2) {
        if (n2 == 0) {
            return n1;
        }
        return gcd(n2, n1 % n2);
    }
}

class BIT16 {

    int[] bit;

    public BIT16(int size) {
        bit = new int[size];
    }

    public void update(int ind, int delta) {
        while (ind < bit.length) {
            bit[ind] += delta;
            ind = ind + (ind & (-1 * ind));
        }
    }

    public int sum(int ind) {
        int s = 0;
        while (ind > 0) {
            s += bit[ind];
            ind = ind - (ind & (-1 * ind));
        }
        return s;
    }

    public int query(int l, int r) {
        return sum(r) - sum(l);
    }
}

class UnionFind18 {

    int[] id;

    public UnionFind18(int size) {
        id = new int[size];
        for (int i = 0; i < size; i++) {
            id[i] = i;
        }
    }

    public int find(int p) {
        int root = p;
        while (root != id[root]) {
            root = id[root];
        }
        while (p != root) {
            int next = id[p];
            id[p] = root;
            p = next;
        }
        return root;
    }

    public void union(int p, int q) {
        int a = find(p), b = find(q);
        if (a == b) {
            return;
        }
        id[b] = a;
    }
}

class Reader {

    final private int BUFFER_SIZE = 1 << 16;
    private DataInputStream din;
    private byte[] buffer;
    private int bufferPointer, bytesRead;

    public Reader() {
        din = new DataInputStream(System.in);
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public Reader(String file_name) throws IOException {
        din = new DataInputStream(new FileInputStream(file_name));
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public String readLine() throws IOException {
        byte[] buf = new byte[64];
        int cnt = 0, c;
        while ((c = read()) != -1) {
            if (c == '\n') {
                break;
            }
            buf[cnt++] = (byte) c;
        }
        return new String(buf, 0, cnt);
    }

    public int ni() throws IOException {
        int ret = 0;
        byte c = read();
        while (c <= ' ') {
            c = read();
        }
        boolean neg = (c == '-');
        if (neg) {
            c = read();
        }
        do {
            ret = ret * 10 + c - '0';
        } while ((c = read()) >= '0' && c <= '9');
        if (neg) {
            return -ret;
        }
        return ret;
    }

    public long nextLong() throws IOException {
        long ret = 0;
        byte c = read();
        while (c <= ' ') {
            c = read();
        }
        boolean neg = (c == '-');
        if (neg) {
            c = read();
        }
        do {
            ret = ret * 10 + c - '0';
        } while ((c = read()) >= '0' && c <= '9');
        if (neg) {
            return -ret;
        }
        return ret;
    }

    public double nextDouble() throws IOException {
        double ret = 0, div = 1;
        byte c = read();
        while (c <= ' ') {
            c = read();
        }
        boolean neg = (c == '-');
        if (neg) {
            c = read();
        }
        do {
            ret = ret * 10 + c - '0';
        } while ((c = read()) >= '0' && c <= '9');
        if (c == '.') {
            while ((c = read()) >= '0' && c <= '9') {
                ret += (c - '0') / (div *= 10);
            }
        }
        if (neg) {
            return -ret;
        }
        return ret;
    }

    private void fillBuffer() throws IOException {
        bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
        if (bytesRead == -1) {
            buffer[0] = -1;
        }
    }

    private byte read() throws IOException {
        if (bufferPointer == bytesRead) {
            fillBuffer();
        }
        return buffer[bufferPointer++];
    }

    public void close() throws IOException {
        if (din == null) {
            return;
        }
        din.close();
    }
}

class FastScanner58 {

    BufferedReader br;
    StringTokenizer st;

    public FastScanner58() {
        br = new BufferedReader(new InputStreamReader(System.in), 32768);
        st = null;
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int ni() {
        return Integer.parseInt(next());
    }

    int[] intArray(int N) {
        int[] ret = new int[N];
        for (int i = 0; i < N; i++) {
            ret[i] = ni();
        }
        return ret;
    }

    long nl() {
        return Long.parseLong(next());
    }

    long[] longArray(int N) {
        long[] ret = new long[N];
        for (int i = 0; i < N; i++) {
            ret[i] = nl();
        }
        return ret;
    }

    double nd() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}