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:
/*
Date: 2017/02/25 19:33:12 Saturday
*/
#include<bits/stdc++.h>
using namespace std;
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define rep(i,a,n) for(int i=(a);i<=(n);++i)
#define dep(i,a,n) for(int i=(a);i>=(n);--i)
#define eps 1e-8
#define pi 3.1415926535897
#define sqr(x) ((x)*(x))
#define MAX(a,b) a=max(a,b)
#define MIN(a,b) a=min(a,b)
#define SZ(x) ((int)(x).size())
#define CPY(a,b) memcpy(a,b,sizeof(a))
#define CLR(a) memset(a,0,sizeof(a))
#define POSIN(x,y) (1<=(x)&&(x)<=n&&1<=(y)&&(y)<=m)
#define all(x) (x).begin(),(x).end()
#define COUT(S,x) cout<<fixed<<setprecision(x)<<S<<endl
#define buli(x) (__builtin_popcountll(x))
#define bur0(x) (__builtin_ctzll(x))
#define bul2(x) (63-__builtin_clzll(x))
#define pw(x) ((ll(1))<<(x))
#define upmo(a,b) (((a)=((a)+(b))%mo)<0?(a)+=mo:(a))
#define mmo(a,b) (((a)=1ll*(a)*(b)%mo)<0?(a)+=mo:(a))
#define y0 y0z
#define y1 y1z
#define yn ynz
#define j0 j0z
#define j1 j1z
#define jn jnz
#define tm tmz
#ifdef LOCAL
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...) 
#endif
typedef long long ll;
typedef double lf;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<lf,lf> pff;
typedef complex<double> CD;
const int inf=0x3f3f3f3f;
const int mo=1000000007;
inline void gn(long long&x){
	int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
	while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;
}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
inline void gn(double&x){double t;scanf("%lf",&t);x=t;}
inline void gn(long double&x){double t;scanf("%lf",&t);x=t;}
template<class T1,class T2>inline void gn(T1&r,T2&s){gn(r),gn(s);}
template<class T1,class T2,class T3>inline void gn(T1&r,T2&s,T3&t){gn(r),gn(s),gn(t);}
template<class T1,class T2,class T3,class T4>inline void gn(T1&r,T2&s,T3&t,T4&u){gn(r),gn(s),gn(t),gn(u);}
inline void gs(char *s){scanf("%s",s);}
inline void gc(char &c){while((c=getchar())>126 || c<33);}
inline void pc(char c){putchar(c);}
const int DX[]={1,0,-1,0},DY[]={0,1,0,-1};
ll powmod(ll a,ll b) {ll res=1;a%=mo;for(;b;b>>=1){if(b&1)res=res*a%mo;a=a*a%mo;}return res;}
ll powmod(ll a,ll b,ll mo) {ll res=1;a%=mo;for(;b;b>>=1){if(b&1)res=res*a%mo;a=a*a%mo;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
//*******************************************

const int N=511111,M=111111;
int l,m,n,t,C;
char s[N];
int dp[N][27],a[N][22];
int get(int x,int y){
	dep(i,20,0)if(a[x][i]&&a[x][i]<=y)x=a[x][i];
	return x==y;
}
int main(){
#ifdef LOCAL
	freopen("C.in","r",stdin);//freopen("C.out","w",stdout);
#endif
	scanf("%s",s+1);n=strlen(s+1);
	dep(i,n,1){
		dp[i][s[i]-'a']=i;
		rep(j,s[i]-'a'+1,26)dp[i][j]=dp[dp[i][j-1]+1][j-1];
		a[i][0]=dp[i][26]==0?n+10:dp[i][26];
		rep(j,0,s[i]-'a'-1)dp[i][j]=dp[a[i][0]+1][j];
		//printf("%d %d\n",i,a[i][0]);
	}
	rep(i,1,n)a[i][0]++;
	rep(j,1,20)rep(i,1,n)a[i][j]=a[a[i][j-1]][j-1];
	scanf("%d",&m);rep(_,1,m){
		int x,y;
		scanf("%d%d",&x,&y);
		puts(get(x,y+1)?"Yes":"No");
	}
	return 0;
}