Problem: 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 ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:

  1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s; 
  2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t; 
  3. 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":

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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 ≤ l ≤ r ≤ |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 ≤ |s| ≤ 2 ⋅ 10^5).

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

Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
template <class A, class B, class C>
struct triple {
  A first;
  B second;
  C third;
  bool operator<(const triple &t) const {
    if (first != t.first) return first < t.first;
    if (second != t.second) return second < t.second;
    return third < t.third;
  }
};
const long double EPS = 0.000000001;
const long double INFD32 = 1000000000;
const long long INF64 = 1000000000000000000;
const long long INF32 = 1000000000;
const long long one = 1;
bool iss_ll(long long __1, long long __2) { return 2 * INF64 / __1 >= __2; }
bool equal(double _1, double _2) { return (abs(_1 - _2) < EPS); }
bool iss_sqr(long long &__1) {
  long long __2 = sqrt((long double)__1);
  return __2 * __2 == __1;
};
bool iss_prime(long long __1) {
  for (long long i = 2; i * i <= __1; ++i)
    if (!(__1 % i)) return 0;
  return 1;
}
long long gcd(long long __1, long long __2) {
  while (__1 && __2)
    if (__1 > __2)
      __1 = __1 % __2;
    else
      __2 = __2 % __1;
  return __1 + __2;
}
long long lcm(long long __1, long long __2) {
  return __1 * __2 / (gcd(__1, __2));
}
long long sq_cube(long long x) {
  if (x == 0) return 1;
  long long l = 0, r = 2e6;
  while (l + 1 < r) {
    long long m = (l + r) / 2;
    if (m * m * m > x)
      r = m;
    else
      l = m;
  }
  return l;
}
long long bp(long long a, long long b, long long md) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    a %= md;
    res %= md;
    b >>= 1;
  }
  return res;
}
vector<long long> z_function(string &s) {
  long long n = ((long long)((s).size()));
  vector<long long> ans(n, 0);
  long long l = 0, r = 0;
  for (long long i = 1; i < n; ++i) {
    ans[i] = min(r - i + 1, ans[i - l]);
    while (ans[i] + i < n && s[ans[i]] == s[i + ans[i]]) {
      ++ans[i];
    }
    if (i + ans[i] - 1 > r) {
      l = i, r = i + ans[i] - 1;
    }
    r = max(r, i);
  }
  return ans;
}
long long arr[500500];
long long cnt[500500][26];
vector<long long> G[26];
long long fl(vector<long long> &v, long long l, long long r) {
  long long pos = 0;
  for (long long i = 20; i >= 0; ++i) {
    if ((pos + (1 << i)) >= v.size()) continue;
    if (v[pos + (1 << i)] < l) pos += (1 << i);
  }
  if (pos == v.size() - 1 || v[pos + 1] > r) return -1;
  return pos + 1;
}
long long fr(vector<long long> &v, long long l, long long r) {
  long long pos = 0;
  for (long long i = 20; i >= 0; ++i) {
    if ((pos + (1 << i)) >= v.size()) continue;
    if (v[pos + (1 << i)] > r) pos += (1 << i);
  }
  if (pos == 0 || v[pos - 1] < l) return -1;
  return pos - 1;
}
void s() {
  string s;
  cin >> s;
  for (long long i = 1; i <= s.size(); ++i) {
    for (long long j = 0; j < 26; ++j) cnt[i][j] = cnt[i - 1][j];
    cnt[i][s[i - 1] - 'a']++;
    G[s[i - 1] - 'a'].push_back(i);
  }
  long long m;
  cin >> m;
  while (m--) {
    long long l, r;
    cin >> l >> r;
    long long len = r - l + 1;
    if (len == 1) {
      cout << "Yes"
           << "\n";
      continue;
    }
    long long cn = 0;
    for (long long i = 0; i < 26; ++i)
      if ((cnt[r][i] - cnt[l - 1][i]) > 0) ++cn;
    if (cn == 1) {
      cout << "No"
           << "\n";
      continue;
    }
    if (s[l - 1] != s[r - 1]) {
      cout << "Yes"
           << "\n";
      continue;
    } else {
      if (cn > 2) {
        cout << "Yes"
             << "\n";
      } else
        cout << "No"
             << "\n";
    }
  }
}
signed main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    s();
  }
  return 0;
}