Problem: You are developing a robot that processes strings. When the robot is given a string t consisting of lowercase English letters, it processes the string by following the procedure below:

1. Let i be the smallest index such that t_i = t_{i + 1}. If such an index does not exist, terminate the procedure.
2. If t_i is `z`, remove t_i and t_{i + 1} from t. Otherwise, let c be the next letter of t_i in the English alphabet, and replace t_i and t_{i + 1} together with c, reducing the length of t by 1.
3. Go back to step 1.



For example, when the robot is given the string `axxxxza`, it will be processed as follows: `axxxxza` → `ayxxza` → `ayyza` → `azza` → `aa` → `b`.

You are given a string s consisting of lowercase English letters. Answer Q queries. The i-th query is as follows:

* Assume that the robot is given a substring of s that runs from the l_i-th character and up to the r_i-th character (inclusive). Will the string be empty after processing?

Constraints

* 1 ≤ |s| ≤ 5 × 10^5
* s consists of lowercase English letters.
* 1 ≤ Q ≤ 10^5
* 1 ≤ l_i ≤ r_i ≤ |s|

Input

The input is given from Standard Input in the following format:


s
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print Q lines. The i-th line should contain the answer to the i-th query: `Yes` or `No`.

Examples

Input

axxxxza
2
1 7
2 6


Output

No
Yes


Input

aabcdefghijklmnopqrstuvwxyz
1
1 27


Output

Yes


Input

yzyyyzyzyyyz
8
1 6
7 12
1 12
6 11
1 1
1 3
4 9
3 8


Output

Yes
Yes
Yes
Yes
No
No
No
No

Solution (in CPP):

Solution:
#include <stdio.h>

char S[500500];
int C[500500][27],V[500500][19];

int main()
{
	scanf ("%s",S+1);
	int L = 0;
	while (S[L+1]) L++;
	for (int i=L;i>=1;i--){
		int p = S[i]-'a';
		C[i][p] = i+1;
		for (int j=p+1;j<=26;j++) C[i][j] = C[C[i][j-1]][j-1];
		V[i][0] = C[i][26];
		if (V[i][0]) for (int j=0;j<p;j++) C[i][j] = C[V[i][0]][j];
		for (int j=1;j<19;j++) V[i][j] = V[V[i][j-1]][j-1];
	}

	int Q; scanf ("%d",&Q); while (Q--){
		int l,r; scanf ("%d %d",&l,&r); r++;

		for (int i=18;i>=0;i--){
			if (V[l][i] && V[l][i] <= r) l = V[l][i];
		}

		puts(l == r? "Yes":"No");
	}
	return 0;
}