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<cstring>
#include<algorithm>
#define rep(i,a,b) for(i=a;i<=b;i++)
using namespace std;
const int N=500010;
const int K=25;
int dp[N][26];
int fa[N][21];
int a[N];char s[N];
int main(){
	//freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
	int i,j,x,n,last,l,r,Q;
	scanf("%s",s+1);n=strlen(s+1);
	rep(i,1,n)	a[i]=s[i]-'a';
	rep(i,1,n+1)	rep(j,0,K)	dp[i][j]=-1;
	rep(i,1,n+1)	fa[i][0]=n+2;
	fa[n+2][0]=n+2;
	for(i=n;i;i--){
		dp[i][a[i]]=i;last=i+1;
		rep(j,a[i]+1,K){
			if(last==n+1)	dp[i][j]=-1;
			else{
				if(dp[last][j-1]!=-1){dp[i][j]=dp[last][j-1];last=dp[i][j]+1;}
				else last=n+1;
			}
		}
		if(dp[i][K]!=-1&&dp[dp[i][K]+1][K]!=-1){
			x=dp[dp[i][K]+1][K]+1;
			rep(j,0,K)	if(dp[i][j]==-1)	dp[i][j]=dp[x][j];
			fa[i][0]=x;
		}
	}
	rep(j,1,20)	rep(i,1,n+2)	fa[i][j]=fa[fa[i][j-1]][j-1];
	scanf("%d",&Q);
	while(Q--){
		scanf("%d%d",&l,&r);r++;
		for(i=20;i>=0;i--)	if(fa[l][i]<=r)	l=fa[l][i];
		if(l==r)	printf("Yes\n");
		else printf("No\n");
	}
}