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 <iostream>
#include <vector>
#include <cstdio>
#include <sstream>
#include <map>
#include <string>
#include <algorithm>
#include <queue>
#include <cmath>
#include <functional>
#include <set>
#include <ctime>
#include <random>
#include <chrono>
#include <cassert>
#include <tuple>
#include <utility>
using namespace std;
 
namespace {
  using Integer = long long; //__int128;
  template<class T, class S> istream& operator >> (istream& is, pair<T,S>& p){return is >> p.first >> p.second;}
  template<class T> istream& operator >> (istream& is, vector<T>& vec){for(T& val: vec) is >> val; return is;}
  template<class T> istream& operator ,  (istream& is, T& val){ return is >> val;}
  template<class T, class S> ostream& operator << (ostream& os, const pair<T,S>& p){return os << p.first << " " << p.second;}
  template<class T> ostream& operator << (ostream& os, const vector<T>& vec){for(size_t i=0; i<vec.size(); i++) os << vec[i] << (i==vec.size()-1?"":" "); return os;}
  template<class T> ostream& operator ,  (ostream& os, const T& val){ return os << " " << val;}
 
  template<class H> void print(const H& head){ cout << head; }
  template<class H, class ... T> void print(const H& head, const T& ... tail){ cout << head << " "; print(tail...); }
  template<class ... T> void println(const T& ... values){ print(values...); cout << endl; }
 
  template<class H> void eprint(const H& head){ cerr << head; }
  template<class H, class ... T> void eprint(const H& head, const T& ... tail){ cerr << head << " "; eprint(tail...); }
  template<class ... T> void eprintln(const T& ... values){ eprint(values...); cerr << endl; }
 
  class range{ Integer start_, end_, step_; public: struct range_iterator{ Integer val, step_; range_iterator(Integer v, Integer step) : val(v), step_(step) {} Integer operator * (){return val;} void operator ++ (){val += step_;} bool operator != (range_iterator& x){return step_ > 0 ? val < x.val : val > x.val;} }; range(Integer len) : start_(0), end_(len), step_(1) {} range(Integer start, Integer end) : start_(start), end_(end), step_(1) {} range(Integer start, Integer end, Integer step) : start_(start), end_(end), step_(step) {} range_iterator begin(){ return range_iterator(start_, step_); } range_iterator   end(){ return range_iterator(  end_, step_); } };
 
  inline string operator "" _s (const char* str, size_t size){ return move(string(str)); }
  constexpr Integer my_pow(Integer x, Integer k, Integer z=1){return k==0 ? z : k==1 ? z*x : (k&1) ? my_pow(x*x,k>>1,z*x) : my_pow(x*x,k>>1,z);}
  constexpr Integer my_pow_mod(Integer x, Integer k, Integer M, Integer z=1){return k==0 ? z%M : k==1 ? z*x%M : (k&1) ? my_pow_mod(x*x%M,k>>1,M,z*x%M) : my_pow_mod(x*x%M,k>>1,M,z);}
  constexpr unsigned long long operator "" _ten (unsigned long long value){ return my_pow(10,value); }
 
  inline int k_bit(Integer x, int k){return (x>>k)&1;} //0-indexed
 
  mt19937 mt(chrono::duration_cast<chrono::nanoseconds>(chrono::steady_clock::now().time_since_epoch()).count());
 
  template<class T> string join(const vector<T>& v, const string& sep){ stringstream ss; for(size_t i=0; i<v.size(); i++){ if(i>0) ss << sep; ss << v[i]; } return ss.str(); }
 
  inline string operator * (string s, int k){ string ret; while(k){ if(k&1) ret += s; s += s; k >>= 1; } return ret; }
}
constexpr long long mod = 9_ten + 7;
constexpr int inf = 114514893;

class closure{
public:
  int unko(){
    for(int i=s.size()-1; i>=0; i--){
      for(int c=26; c>=0; c--){
        if( c == s[i]-'a'){
          dp[ c ][ i ] = i+1;
        }else if(c < s[i]-'a'){
          if(dp[26][i] != inf) dp[ c ][ i ] = dp[ c ][ dp[ 26 ][ i ] ];
        }else{
          int r = i+1;
          for(int cc=s[i]-'a'; cc<c; cc++){
            if(r == inf) break;
            r = dp[ cc ][ r ];
          }
          dp[ c ][ i ] = r;
        }
      }
    }
  }

  string s;
  vector<vector<int>> dp;
  closure(string s) : s(s), dp(27,vector<int>(s.size()+1, -1)) {
    for(int i=0; i<s.size(); i++){
      dp[ s[i]-'a' ][ i ] = i+1;
    }
    for(int c=0; c<27; c++){
      dp[ c ][ s.size() ] = inf;
    }

    // unko();
  }


  int g(int c, int i){
    if( i >= s.size()) return inf;
    if( dp[c][i] != -1 ) return dp[c][i];

    if( c+'a' < s[i] ){
      int r = i+1;
      for(int cc=s[i]-'a'; cc<26; cc++){
        r = g(cc, r);
      }
      return dp[c][i] = g(c, r);
    }else{
      int r = i+1;
      for(int cc=s[i]-'a'; cc<c; cc++){
        r = g(cc, r);
        if( r == inf ) return dp[c][i] = inf;
      }
      return dp[c][i] = r;
    }
  }

  int operator() (int i){
    return g(26,i);
  }

};

int main(){
  string s;
  cin >> s;

  closure g(s);

  vector<vector<int>> emp(s.size()+1, vector<int>(21, inf));
  for(int i=s.size()-1; i>=0; i--){
    for(int c=26; c>=0; c--){
      g.g(c,i);
    }
    emp[ i ][ 0 ] = g(i);
  }

  for(int k=1; k<21; k++){
    for(int i=0; i+(1<<k)<s.size(); i++){
      if( emp[ i ][ k-1 ] == inf ) emp[i][k] = inf;
      else emp[ i ][ k ] = emp[ emp[ i ][ k-1 ] ][ k-1 ];
    }
  }

  // return 0;

  int q;
  cin >> q;

  while(q--){
    int l,r;
    // cin >> l,r;
    scanf("%d%d", &l, &r);
    l--;

    if(r-l == 1){
      println("No");
      continue;
    }

    int k = 20;
    while( k>=0 && l<r ){
      if( emp[ l ][ k ] <= r ){
        l = emp[l][k];
      }
      k--;
    }

    // while( l< r){
    //   l = g(l);
    // }

    if( l==r ){
      println("Yes");
    }else{
      println("No");
    }
  }

  return 0;
}