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:
#define DEB
#include<bits/stdc++.h>
#define REP(i,m) for(int i=0;i<(m);++i)
#define REPN(i,m,in) for(int i=(in);i<(m);++i)
#define ALL(t) (t).begin(),(t).end()
#define CLR(a) memset((a),0,sizeof(a))
#define pb push_back
#define mp make_pair
#define fr first
#define sc second

using namespace std;


#ifdef DEB
#define dump(x)  cerr << #x << " = " << (x) << endl
#define prl cerr<<"called:"<< __LINE__<<endl
#define dumpR(x) cerr<<"\x1b[31m"<<#x<<" = " <<(x)<<"\x1b[39m"<<endl
#define dumpY(x) cerr<<"\x1b[33m"<<#x<<" = " <<(x)<<"\x1b[39m"<<endl
#define dumpG(x) cerr<<"\x1b[32m"<<#x<<" = " <<(x)<<"\x1b[39m"<<endl
template<class T> void debug(T a,T b){ for(;a!=b;++a) cerr<<*a<<' ';cerr<<endl;}
#else
#define dump(x) ;
#define dumpR(x) ;
#define dumpY(x) ;
#define dumpG(x) ;
#define prl ;
template<class T> void debug(T a,T b){ ;}
#endif

template<class T> void chmin(T& a,const T& b) { if(a>b) a=b; }
template<class T> void chmax(T& a,const T& b) { if(a<b) a=b; }

typedef long long int lint;
typedef pair<int,int> pi;

namespace std{
  template<class S,class T>
  ostream &operator <<(ostream& out,const pair<S,T>& a){
    out<<'('<<a.fr<<','<<a.sc<<')';
    return out;
  }
}

const int INF=5e8;

char s[500005];
int q,n;

int jump[27][500005];

int blank[20][500005];
int main(){
  scanf("%s",s);
  n=strlen(s);

  REP(c,26) jump[c][n]=INF;
  jump[26][n]=n;
  for(int i=n-1;i>=0;--i){
    int c2=s[i]-'a';
    REPN(c,27,c2){
      if(s[i]=='a'+c) jump[c][i]=i+1;
      else{
        if(c==0 || jump[c-1][i]==INF) jump[c][i]=INF;
        else jump[c][i]=jump[c-1][jump[c-1][i]];
      }
    }
    REP(c,c2){
      if(jump[26][i]<INF) jump[c][i]=jump[c][jump[26][i]];
      else jump[c][i]=INF;
    }
  }
  REP(i,n) blank[0][i]=jump[26][i];
  REP(i,20) blank[i][n]=n;
  REP(i,19) REP(j,n){
    if(blank[i][j]<INF) blank[i+1][j]=blank[i][blank[i][j]];
    else blank[i+1][j]=INF;
  }

  cin>>q;
  while(q--){
    int l,r;scanf("%d%d",&l,&r);--l;
    for(int i=19;i>=0;--i) if(blank[i][l]<=r){
      l=blank[i][l];
    }
    if(l==r) puts("Yes");
    else puts("No");
  }
  return 0;
}