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<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<cmath>
#include<string>

#define ls (t<<1)
#define rs ((t<<1)+1)
#define mid ((l+r)>>1)
#define fi first
#define se second
#define mk make_pair
#define pb push_back

#define N 500005
#define K 19
#define M 200005
#define seed 23333

using namespace std;
const int inf=(int)1e9;
int i,j,m,n,p,k,f[N][27],B[N][K+1],Q,l,r;
char s[N];
int main()
{
	scanf("%s",s+1);
	n=strlen(s+1);
	memset(f,60,sizeof(f));
	for (i=n;i;--i)
	{
			f[i][s[i]-'a']=i+1;
				for (j=s[i]-'a'+1;j<=s[i]-'a'+26;++j)
			{
					int now=j%27,la=(j+26)%27;
					if (j>=27) la=26;
					if (f[i][la]>inf) continue;
					if (j>=27)
					f[i][now]=f[f[i][la]][now];
					else 
					f[i][now]=f[f[i][la]][la];
			}
	}
	memset(B,60,sizeof(B));
	for (i=n;i;--i)
	{
			B[i][0]=f[i][26];
			for (j=1;j<=K;++j)
			{
					if (B[i][j-1]>inf) B[i][j]=inf+5;
					else B[i][j]=B[B[i][j-1]][j-1];
			}
	}
	scanf("%d",&Q);
	for (;Q--;)
	{
			scanf("%d%d",&l,&r);
			for (i=K;i>=0;--i)
				if (B[l][i]<=r+1) l=B[l][i];
			if (l==r+1) puts("Yes");
			else puts("No"); 
	}
}