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;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const int MN = 500010;

int tor[MN][27]; //26 for ""
int nx[20][MN];

int main() {
	string s;
	cin >> s;
	int n = s.size();

	rep(i, 27) tor[n][i] = MN;

	for (int i = n-1; i >= 0; --i) {
		int c = s[i] - 'a';
		tor[i][c] = i + 1;

		for (int j = c + 1; j < 27; ++j) {
			tor[i][j] = (tor[i][j-1] == MN ? MN : tor[tor[i][j-1]][j-1]);
		}

		nx[0][i] = tor[i][26];

		for (int j = 0; j < c; ++j) {
			tor[i][j] = (tor[i][26] == MN ? MN : tor[tor[i][26]][j]);
		}
	}

	rep(i, 20) nx[i][n] = MN;

	rep(i, 19) {
		rep(j, n) {
			nx[i+1][j] = (nx[i][j] == MN ? MN : nx[i][nx[i][j]]);
		}
	}

	int q;
	cin >> q;
	while (q--) {
		int l, r;
		cin >> l >> r;
		--l;

		for (int i = 19; i >= 0; --i) {
			if (nx[i][l] <= r) {
				l = nx[i][l];
			}
		}

		puts(l == r ? "Yes" : "No");
	}

	return 0;
}