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:
// <string.cpp> - Sat Mar 25 20:06:07 2017
// This file is created by XuYike's black technology automatically.
// Copyright (C) 2015 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long lol;
template<typename T>
inline void gg(T &res){
    res=0;T fh=1;char ch=getchar();
    while((ch>'9'||ch<'0')&&ch!='-')ch=getchar();
    if(ch=='-')fh=-1,ch=getchar();
    while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
    res*=fh;
}
inline int gi(){int x;gg(x);return x;}
inline lol gl(){lol x;gg(x);return x;}
const int MAXN=500010;
const int K=20;
const int INF=1e9;
char s[MAXN];
int a[MAXN][27],b[K][MAXN];
int main(){
    scanf("%s",s+1);
    int n=strlen(s+1);
    for(int i=n;i;i--){
        a[i][s[i]-'a']=i+1;
        for(int j=s[i]-'a'+1;j<=26;j++)a[i][j]=a[a[i][j-1]][j-1];
        for(int j=0;j<s[i]-'a';j++)a[i][j]=a[a[i][26]][j];
    }
    for(int i=1;i<=n;i++)b[0][i]=a[i][26];
    for(int j=1;j<K;j++)
        for(int i=1;i<=n;i++)b[j][i]=b[j-1][b[j-1][i]];
    int q=gi();
    while(q--){
        int l=gi(),r=gi()+1;
        for(int i=K-1;i>=0;i--)
            if(b[i][l]&&b[i][l]<=r)l=b[i][l];
        puts(l==r?"Yes":"No");
    }
    return 0;
}