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 <bits/stdc++.h>

using namespace std;

using ll = long long;
using vi = vector<int>;
using db = double;
#define pb push_back

// 看了题解，发现当时想不出的原因主要不是倍长，而是倍长的前一步的DP没想清楚

const int mxn = 500006;
const int emp = 26;

char a[mxn];
int n;
int f[mxn][27]; // 无效均为0
int s[mxn][20];

int main() {
	scanf("%s", a + 1);
	n = strlen(a + 1);
	for (int i = 1; i <= n; ++i)
		a[i] -= 'a';
	
	for (int i = n; i >= 1; --i) {
		f[i][a[i]] = i + 1;
		for (int j = a[i] + 1; j <= 26; ++j)
			f[i][j] = f[f[i][j - 1]][j - 1];
		for (int j = 0; j < a[i]; ++j)
			f[i][j] = f[f[i][emp]][j];
	}
	
	for (int i = n; i >= 1; --i) {
		s[i][0] = f[i][emp];
		for (int j = 1; j < 20; ++j)
			s[i][j] = s[s[i][j - 1]][j - 1];
	}
	
	int q;
	cin >> q;
	while (q--) {
		int l, r;
		scanf("%d%d", &l, &r);
		++r;
		for (int i = 20; i--; ) if (s[l][i] && s[l][i] <= r)
			l = s[l][i];
		puts(l == r ? "Yes" : "No");
	}
	
	return 0;
}