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:
// srt: 2017年02月28日 15時26分48秒
// 《别老母》  黄景仁
// 搴帷拜母河梁去，白发愁看泪眼枯。
// 惨惨柴门风雪夜，此时有子不如无。
// end: 2017年02月28日 15時55分06秒

#include<cstdio>
#include<cstring>
inline void read(int&r)
{
	register char c;r=0;
	do c=getchar();while(c<'0'||c>'9');
	do r=r*10+c-'0',c=getchar();while(c>='0'&&c<='9');
}

char s[500001];
int next[27][500002];
#define min(a,b) a<b?a:b

int main()
{
	scanf("%s",s);

	int i,k,c,n=strlen(s),inf=n+1;
	for(k=0;k<=26;k++)next[k][n]=next[k][inf]=inf;
	for(i=n-1;i>=0;i--)
	{
		c=s[i]-'a'+1;
		next[c][i]=i+1;
		for(k=c+1;k<=26;k++)
			next[k][i]=next[k-1][next[k-1][i]];
		next[0][i]=next[26][next[26][i]];
		if(c>1)next[1][i]=next[1][next[0][i]];
		for(k=2;k<c;k++)
			next[k][i]=min(next[k-1][next[k-1][i]],next[k][next[0][i]]);
	}
	for(k=1;1<<k<=n;k++)
		for(i=0;i<n;i++)
			next[k][i]=next[k-1][next[k-1][i]];
	c=k-1;

	int q,l,r;read(q);
	while(q--)
	{
		read(l),read(r),l--;
		for(k=c;k>=0;k--)
			if(next[k][l]<=r)
				l=next[k][l];
		puts(l==r?"Yes":"No");
	}
}