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 <sstream>
#include <cassert>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <string>
#include <queue>
#include <bitset>
using namespace std;

#define int long long
#define FOR(i, a, b) for (int i = (a), _b = (b); i <= _b; ++i)
#define FORD(i, a, b) for (int i = (a), _b = (b); i >= _b; --i)
#define REP(i, a) for (int i = 0, _a = (a); i < _a; ++i)
#define REPD(i,n) for(int i = (n)-1; i >= 0; --i)

#define DEBUG(X) { cerr << #X << " = " << (X) << endl; }
#define PR(A, n) { cerr << #A << " = "; FOR(_, 1, n) cerr << A[_] << ' '; cerr << endl; }
#define PR0(A, n) { cerr << #A << " = "; REP(_, n) cerr << A[_] << ' '; cerr << endl; }

#define sqr(x) ((x) * (x))
#define ll long long
#define double long double
typedef pair<int, int> II;
#define PI (2 * acos((double)0))
#define __builtin_popcount __builtin_popcountll
#define SZ(x) ((int)(x).size())
#define ALL(a) (a).begin(), (a).end()
#define MS(a,x) memset(a, x, sizeof(a))
#define next ackjalscjaowjico
#define prev ajcsoua0wucckjsl
#define y1 alkscj9u20cjeijc
#define y0 u9cqu3jioajc

double safe_sqrt(double x) { return sqrt(max((double)0.0, x)); }
int GI(int& x) { return scanf("%lld", &x); }

const int MN = 500111;
string s;
int a[MN];
int f[MN][33], g[22][MN];

int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout << (fixed) << setprecision(9);
    while (cin >> s) {
        int Z = 'z' - 'a';

        int n = SZ(s);
        FOR(i,1,n) a[i] = s[i-1] - 'a';

        MS(f, 0x3f);
        MS(g, 0x3f);

        FORD(i,n,1) {
            FOR(c,a[i],Z+1) {
                if (c == a[i]) f[i][c] = i;
                else {
                    int j = f[i][c-1];
                    if (j > n) f[i][c] = n+1;
                    else f[i][c] = f[j+1][c-1];
                }
            }

            g[0][i] = f[i][Z + 1];

            FOR(c,0,a[i]-1) {
                int j = g[0][i];

                if (j > n) f[i][c] = n+1;
                else f[i][c] = f[j+1][c];
            }
        }

        FOR(t,1,20) {
            FOR(i,1,n) {
                int j = g[t-1][i];
                if (j > n) g[t][i] = n+1;
                else g[t][i] = g[t-1][j+1];
            }
        }

        int q; cin >> q;
        REP(qq,q) {
            int l, r; cin >> l >> r;
            ++r;
            FORD(t,20,0) {
                if (g[t][l] < r) {
                    l = g[t][l] + 1;
                    if (l == r) break;
                }
            }
            cout << (l == r ? "Yes" : "No") << '\n';
        }
    }
}