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

/**
 *
 * @author Saju
 *
 */

public class Main {

	private static int dx[] = { -2, -3, -2, -1, -1, 1};
	private static int dy[] = { 1, -1, -1, -2, -3, -2};

	private static final long INF = (long) (1e15);
	private static final int INT_INF = Integer.MAX_VALUE;
	private static final long NEG_INF = Long.MIN_VALUE;
	private static final double EPSILON = 1e-10;

	private static final int MAX = 1007;
	private static final long MOD = 1000000007;
	
	private static final int MAXN = 100007;
	private static final int MAXA = 10000009;
	private static final int MAXLOG = 22;

	public static void main(String[] args) {

		InputReader in = new InputReader(System.in);
		PrintWriter out = new PrintWriter(System.out);
		

/*


*/
		
		char[] str = in.next().toCharArray();
		int len = str.length;
		
		long[][] sum = new long[len][26];
		
		int val = str[0] - 'a';
		sum[0][val]++;
		
		for(int i = 1; i < len; i++) {
			val = str[i] - 'a';
			for(int j = 0; j < 26; j++) {
				sum[i][j] = sum[i - 1][j];
			}
			sum[i][val] += 1;
			
		}
		
		
		int q = in.nextInt();
		for(int i = 0; i < q; i++) {
			int	 l = in.nextInt();
			int r = in.nextInt();
			l--;
			r--;
			
			if(l == r) {
				out.println("Yes");
			}
			else if(str[l] != str[r]) {
				out.println("Yes");
			}
			else {
			
				int mismatch = 0;
				
				for(int j = 0; j < 26; j++) {
					if(sum[l][j] != sum[r][j]) {
						mismatch++;
					}
				}
				if(mismatch >= 3) {
					out.println("Yes");
				}
				else {
					out.println("No");
				}
				
			}
			
		}
		
		out.flush();
		out.close();
		System.exit(0);
	}
	
	// O(log(max(A,M)))
	static long modInverse(long a, long m) {
		extendedEuclid(a, m);
		return (x % m + m) % m;
	}

	static long x;
	static long y;
	static long gcdx;

	private static void extendedEuclid(long a, long b) {
		if (b == 0) {
			gcdx = a;
			x = 1;
			y = 0;
		} else {
			extendedEuclid(b, a % b);
			long temp = x;
			x = y;
			y = temp - ((a / b) * y);
		}
	}
	
	private static void generatePrime(int n) {
		 //O(NloglogN)
		boolean arr[] = new boolean[n + 5];
		Arrays.fill(arr, true);
		for(int i = 2; i * i <=n; i++){
			if(arr[i] == true){
				for(int j = i * i; j <= n; j+=i){
					arr[j] = false; 
				}
			}
		}
		int count = 0;
		int start = 0;
		for(int i = 2; i <= n; i++){
			if(arr[i] == true){
//				System.out.println(i + " ");
				count++;
			}
			if(count == (start * 100) + 1) {
//				System.out.println(i);
				start++;
			}
		}
		System.out.println();
		System.out.println(count);
		
	}
	
	private static Map<Long, Long> primeFactorization(long n, long m){
		Map<Long, Long> map = new HashMap<>();
		
		for(long i = 2; i <= Math.sqrt(n); i++){
			if(n % i == 0){
			long count = 0;
			while(n % i == 0){
				count++;
				n = n / i;
			}
			long val = count * m;
			map.put(i, val);
//			System.out.println("i: " + i + ", count: " + count);
		}
			
		}
		if(n != 1) {
//			System.out.println(n);
			map.put(n, m);
		}
		return map;
	}

	private static class Pair<T>{
		T a;
		T b;
		Pair(T a, T b){
			this.a = a;
			this.b = b;
		}
	}
	
	private static class SegmentTree{
		int n;
		private final int MAXN = 100007;
		private long[] tree = new long[MAXN << 2];
		private long[] lazy = new long[MAXN << 2];
		private long[] arr = new long[MAXN];
		
		
		/***
		 * 
		 * arr is 1 based index.
		 * 
		 * @param arr
		 * 
		 */
		SegmentTree(int n, long[] arr){
			this.n = n;
			for(int i = 0; i < arr.length; i++) {
				this.arr[i] = arr[i];
			}
		}
		
		void build(int index, int left, int right) {
			if(left == right) {
				tree[index] = arr[left];
			}
			else {
				int mid = (left + right) / 2;
				
				build(index * 2, left, mid);
				build((index * 2) + 1, mid + 1, right);
				tree[index] = max(tree[(index * 2)], tree[(index * 2) + 1]);
			}
		}
		
		long query(int node, int left, int right, int start, int end) {
			
			if(left > end || right < start) {
				return NEG_INF;
			}
			
			if(left >= start && right <= end) {
				return tree[node];
			}
			int mid = (left + right) / 2;
			long val1 = query(2 * node, left, mid, start, end);
			long val2 = query(2 * node + 1, mid + 1, right, start, end);
			
			return max(val1, val2);
		}
		
		void update(int node, int left, int right, int idx, long val) {
			if(left == right) {
				tree[node] += val;
			}
			else {
				int mid = (left + right) / 2;
				if(idx <= mid) {
					update(2 * node, left, mid, idx, val);
				}
				else {
					update(2 * node + 1, mid + 1, right, idx, val);
				}
				tree[node] = max(tree[(2 * node) + 1], tree[(2 * node)]);
			}
		}
		
		

		 void updateRange(int node, int start, int end, int l, int r, long val)
		{
		    if(lazy[node] != 0)
		    { 
		        // This node needs to be updated
		        tree[node] =  lazy[node];    // Update it
		        if(start != end)
		        {
		            lazy[node*2] = lazy[node];                  // Mark child as lazy
		            lazy[node*2+1] = lazy[node];                // Mark child as lazy
		        }
		        lazy[node] = 0;                                  // Reset it
		    }
		    if(start > end || start > r || end < l)              // Current segment is not within range [l, r]
		        return;
		    if(start >= l && end <= r)
		    {
		        // Segment is fully within range
		        tree[node] =  val;
		        if(start != end)
		        {
		            // Not leaf node
		            lazy[node*2] = val;
		            lazy[node*2+1] = val;
		        }
		        return;
		    }
		    int mid = (start + end) / 2;
		    updateRange(node*2, start, mid, l, r, val);        // Updating left child
		    updateRange(node*2 + 1, mid + 1, end, l, r, val);   // Updating right child
		    tree[node] = max(tree[node*2], tree[node*2+1]);        // Updating root with max value 
		}

		 long queryRange(int node, int start, int end, int l, int r)
		{
		    if(start > end || start > r || end < l)
		        return 0;         // Out of range
		    if(lazy[node] != 0)
		    {
		        // This node needs to be updated
		        tree[node] =  lazy[node];            // Update it
		        if(start != end)
		        {
		            lazy[node*2] = lazy[node];         // Mark child as lazy
		            lazy[node*2+1] = lazy[node];    // Mark child as lazy
		        }
		        lazy[node] = 0;                 // Reset it
		    }
		    if(start >= l && end <= r)             // Current segment is totally within range [l, r]
		        return tree[node];
		    int mid = (start + end) / 2;
		    long p1 = queryRange(node*2, start, mid, l, r);         // Query left child
		    long p2 = queryRange(node*2 + 1, mid + 1, end, l, r); // Query right child
		    return max(p1, p2);
		}
		
		 void buildRange(int node, int low, int high){
			if(low == high){
				tree[node] =  arr[low];
				return;
			}
			
			int mid = (low + high) / 2;
			int left = node << 1;
			int right = left | 1;
			buildRange(left, low, mid);
			buildRange(right, mid + 1, high);
			tree[node] = max(tree[left], tree[right]);
		}
		
		
		void printSegmentTree() {
			System.out.println(Arrays.toString(tree));
		}
		
	}
	
	private static class KMP{
		
		private static char[] t;
		private static char[] s;
		
		public int kmp(char[]t, char[]s) {
			this.t = t;
			this.s = s;
			return this.kmp();
		}
		
		private int kmp() {
			
			List<Integer> prefixTable = getPrefixTable(s);
			
			int match = 0;
			int i = 0;
			int j = 0;
			
			int n = t.length;
			int m = s.length;
			while(i < n) {
				if(t[i] == s[j]) {
					if(j == m - 1) {
						match++;
						j = prefixTable.get(j - 1);
						continue;
					}
					i++;
					j++;
				}
				else if(j > 0) {
					j = prefixTable.get(j - 1);
				}
				else {
					i++;
				}
			}
			
			return match;
		}

		/***
		1. We compute the prefix values π[i] in a loop by iterating 	<br/>
				from i=1 to i=n−1 (π[0] just gets assigned with 0).		<br/><br/>
		2. To calculate the current value π[i] we set the variable j 	<br/>
				denoting the length of the best suffix for i−1. Initially j=π[i−1].	<br/>
		3. Test if the suffix of length j+1 is also a prefix by		 <br/><br/>
				comparing s[j] and s[i]. If they are equal then we assign π[i]=j+1, 	<br/>
				otherwise we reduce j to π[j−1] and repeat this step.		<br/><br/>
		4. If we have reached the length j=0 and still don't have a match,	<br/> 
				then we assign π[i]=0 and go to the next index i+1.		<br/><br/>
				
		@param
	 	pattern(String)
	***/
		private List<Integer> getPrefixTable(char[] pattern){
			
			List<Integer> prefixTable = new ArrayList<Integer>();
			int n = pattern.length;
			for(int i = 0; i < n; i++) {
				prefixTable.add(0);
			}
			
			for(int i = 1; i < n; i++) {
				for(int j = prefixTable.get(i - 1); j >= 0;) {
					if(pattern[j] == pattern[i]) {
						prefixTable.set(i, j + 1);
						break;
					}
					else if(j > 0){
						j = prefixTable.get(j - 1);
					}
					else {
						break;
					}
				}
			}
			
			return prefixTable;
		}
	}
	


	private static class Point {
		int x;
		int y;
		Point(int x, int y){
			this.x = x;
			this.y = y;
		}
		
		@Override
		public boolean equals(Object obj) {
			Point ob = (Point) obj;
			if(this.x == ob.x && this.y == ob.y){
				return true;
			}
			return false;
			
		}
		@Override
		public String toString() {
			return this.x + " " + this.y;
		}
		
		@Override
		public int hashCode() {
		    return 0;
		}
	}

	private static long pow(int base, int pow) {
		long val = 1L;
		for(int i = 1; i <= pow; i++) {
			val *= base;
		}
		return val;
	}
	
	private static int log(int x, int base) {
		return (int) (Math.log(x) / Math.log(base));
	}
	private static int log(long x, int base) {
		return (int) (Math.log(x) / Math.log(base));
	}
	private static long max(long a, long b) {
		if (a >= b) {
			return a;
		}
		return b;
	}

	private static long abs(long a) {
		if (a < 0) {
			return -a;
		}
		return a;
	}

	private static int abs(int a) {
		if (a < 0) {
			return -a;
		}
		return a;
	}

	private static int max(int a, int b) {
		if (a >= b) {
			return a;
		}
		return b;
	}

	private static int min(int a, int b) {
		if (a <= b) {
			return a;
		}
		return b;
	}
	
	private static long min(long a, long b) {
		if (a <= b) {
			return a;
		}
		return b;
	}

	private static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	private static int gcd(int a, int b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	private static long bigMod(long n, long k, long m) {

		long ans = 1;
		while (k > 0) {
			if ((k & 1) == 1) {
				ans = (ans * n) % m;
			}
			n = (n * n) % m;
			k >>= 1;
		}
		return ans;
	}

	/*
	 * Returns an iterator pointing to the first element in the range [first,
	 * last] which does not compare less than val.
	 * 
	 */
	private static int lowerBoundNew(long[] arr, long num) {
		int start = 0;
		int end = arr.length - 1;
		int index = 0;
		int len = arr.length;
		int mid = 0;
		while (true) {
			if (start > end) {
				break;
			}
			mid = (start + end) / 2;
			if (arr[mid] > num) {
				end = mid - 1;
			} else if (arr[mid] < num) {
				start = mid + 1;
			} else {
				while (mid >= 0 && arr[mid] == num) {
					mid--;
				}
				return mid + 1;
			}
		}
		if (arr[mid] < num) {
			return mid + 1;
		}
		return mid;
	}

	/*
	 * upper_bound() is a standard library function in C++ defined in the header
	 * . It returns an iterator pointing to the first element in the range
	 * [first, last) that is greater than value, or last if no such element is
	 * found
	 * 
	 */
	private static int upperBoundNew(long[] arr, long num) {

		int start = 0;
		int end = arr.length - 1;
		int index = 0;
		int len = arr.length;
		int mid = 0;
		while (true) {
			if (start > end) {
				break;
			}
			mid = (start + end) / 2;
			if (arr[mid] > num) {
				end = mid - 1;
			} else if (arr[mid] < num) {
				start = mid + 1;
			} else {
				while (mid < len && arr[mid] == num) {
					mid++;
				}
				if (mid == len - 1 && arr[mid] == num) {
					return mid + 1;
				} else {
					return mid;
				}
			}
		}
		if (arr[mid] < num) {
			return mid + 1;
		}
		return mid;
	}

	private static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}

		public String next() {
			try {
				while (tokenizer == null || !tokenizer.hasMoreTokens()) {
					tokenizer = new StringTokenizer(reader.readLine());

				}
			} catch (IOException e) {
				return null;
			}
			return tokenizer.nextToken();
		}

		public String nextLine() {
			String line = null;
			try {
				tokenizer = null;
				line = reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
			return line;
		}

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

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public boolean hasNext() {
			try {
				while (tokenizer == null || !tokenizer.hasMoreTokens()) {
					tokenizer = new StringTokenizer(reader.readLine());
				}
			} catch (Exception e) {
				return false;
			}
			return true;
		}
	}
}