Let's call two strings $s$ and $t$ anagrams of each other if it is possible to rearrange symbols in the string $s$ to get a string, equal to $t$.

Let's consider two strings $s$ and $t$ which are anagrams of each other. We say that $t$ is a reducible anagram of $s$ if there exists an integer $k \ge 2$ and $2k$ non-empty strings $s_1, t_1, s_2, t_2, \dots, s_k, t_k$ that satisfy the following conditions:

  If we write the strings $s_1, s_2, \dots, s_k$ in order, the resulting string will be equal to $s$;  If we write the strings $t_1, t_2, \dots, t_k$ in order, the resulting string will be equal to $t$;  For all integers $i$ between $1$ and $k$ inclusive, $s_i$ and $t_i$ are anagrams of each other. 

If such strings don't exist, then $t$ is said to be an irreducible anagram of $s$. Note that these notions are only defined when $s$ and $t$ are anagrams of each other.

For example, consider the string $s = $ "gamegame". Then the string $t = $ "megamage" is a reducible anagram of $s$, we may choose for example $s_1 = $ "game", $s_2 = $ "gam", $s_3 = $ "e" and $t_1 = $ "mega", $t_2 = $ "mag", $t_3 = $ "e":

 [Image] 

On the other hand, we can prove that $t = $ "memegaga" is an irreducible anagram of $s$.

You will be given a string $s$ and $q$ queries, represented by two integers $1 \le l \le r \le |s|$ (where $|s|$ is equal to the length of the string $s$). For each query, you should find if the substring of $s$ formed by characters from the $l$-th to the $r$-th has at least one irreducible anagram.


-----Input-----

The first line contains a string $s$, consisting of lowercase English characters ($1 \le |s| \le 2 \cdot 10^5$).

The second line contains a single integer $q$ ($1 \le q \le 10^5$)  — the number of queries.

Each of the following $q$ lines contain two integers $l$ and $r$ ($1 \le l \le r \le |s|$), representing a query for the substring of $s$ formed by characters from the $l$-th to the $r$-th.


-----Output-----

For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.


-----Examples-----
Input
aaaaa
3
1 1
2 4
5 5

Output
Yes
No
Yes

Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5

Output
No
Yes
Yes
Yes
No
No



-----Note-----

In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose $s_1 = $ "a", $s_2 = $ "aa", $t_1 = $ "a", $t_2 = $ "aa" to show that it is a reducible anagram.

In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import sys\ninput = sys.stdin.readline\n\nfrom collections import Counter\n\nS=input().strip()\ncount=[[0]*26 for i in range(len(S)+1)]\nc=1\nfor s in S:\n    for k in range(26):\n        count[c][k]=count[c-1][k]\n    count[c][ord(s)-97]+=1\n    c+=1\n    \nQ=int(input())\nfor tests in range(Q):\n    l,r=list(map(int,input().split()))\n    if l==r:\n        print("Yes")\n        continue\n    \n    elif S[l-1]!=S[r-1]:\n        print("Yes")\n        continue\n\n    elif len(S)<=4:\n        print("No")\n        continue\n\n    X=[count[r][j]-count[l-1][j] for j in range(26)]\n    NZERO=0\n    for x in X:\n        if x!=0:\n            NZERO+=1\n\n    if NZERO<=2:\n        print("No")\n    else:\n        print("Yes")\n\n    \n        \n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14147