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":

<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 ≤ 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;
using ll = long long;
using ld = long double;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vpi = vector<pi>;
using si = set<int>;
using sl = set<ll>;
template <class T>
bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
namespace input {
template <class T>
void re(complex<T>& x);
template <class T1, class T2>
void re(pair<T1, T2>& p);
template <class T>
void re(vector<T>& a);
template <class T, size_t SZ>
void re(array<T, SZ>& a);
template <class T>
void re(T& x) {
  cin >> x;
}
void re(double& x) {
  string t;
  re(t);
  x = stod(t);
}
void re(ld& x) {
  string t;
  re(t);
  x = stold(t);
}
template <class T, class... Ts>
void re(T& t, Ts&... ts) {
  re(t);
  re(ts...);
}
template <class T>
void re(complex<T>& x) {
  T a, b;
  re(a, b);
  x = cd(a, b);
}
template <class T1, class T2>
void re(pair<T1, T2>& p) {
  re(p.first, p.second);
}
template <class T>
void re(vector<T>& a) {
  for (int i = 0; i < (((int)(a).size())); ++i) re(a[i]);
}
template <class T, size_t SZ>
void re(array<T, SZ>& a) {
  for (int i = 0; i < (SZ); ++i) re(a[i]);
}
}  // namespace input
using namespace input;
namespace output {
void pr(int x) { cout << x; }
void pr(long x) { cout << x; }
void pr(ll x) { cout << x; }
void pr(unsigned x) { cout << x; }
void pr(unsigned long x) { cout << x; }
void pr(unsigned long long x) { cout << x; }
void pr(float x) { cout << x; }
void pr(double x) { cout << x; }
void pr(ld x) { cout << x; }
void pr(char x) { cout << x; }
void pr(const char* x) { cout << x; }
void pr(const string& x) { cout << x; }
void pr(bool x) { pr(x ? "true" : "false"); }
template <class T>
void pr(const complex<T>& x) {
  cout << x;
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T>
void pr(const T& x);
template <class T, class... Ts>
void pr(const T& t, const Ts&... ts) {
  pr(t);
  pr(ts...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr("{", x.first, ", ", x.second, "}");
}
template <class T>
void pr(const T& x) {
  pr("{");
  bool fst = 1;
  for (const auto& a : x) pr(!fst ? ", " : "", a), fst = 0;
  pr("}");
}
void ps() { pr("\n"); }
template <class T, class... Ts>
void ps(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pr(" ");
  ps(ts...);
}
void pc() { pr("]\n"); }
template <class T, class... Ts>
void pc(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pr(", ");
  pc(ts...);
}
}  // namespace output
using namespace output;
inline int bs(bool (*valid)(int), int l, int r, bool order) {
  l--, r++;
  if (!order) swap(l, r);
  while (abs(r - l) > 1) {
    int mid = (l + r) >> 1;
    if (valid(mid))
      r = mid;
    else
      l = mid;
  }
  valid(r);
  return r;
}
const int M = 1e9 + 7;
const ll lnf = 1e18 + 3;
const int inf = 1e9 + 3;
const int nax = 2e5 + 5;
typedef decay<decltype(M)>::type T;
struct mi {
  T val;
  explicit operator T() const { return val; }
  mi() { val = 0; }
  mi(const ll& v) {
    val = (-M <= v && v <= M) ? v : v % M;
    if (val < 0) val += M;
  }
  friend void pr(const mi& a) { pr(a.val); }
  friend void re(mi& a) {
    ll x;
    re(x);
    a = mi(x);
  }
  friend bool operator==(const mi& a, const mi& b) { return a.val == b.val; }
  friend bool operator!=(const mi& a, const mi& b) { return !(a == b); }
  friend bool operator<(const mi& a, const mi& b) { return a.val < b.val; }
  mi operator-() const { return mi(-val); }
  mi& operator+=(const mi& m) {
    if ((val += m.val) >= M) val -= M;
    return *this;
  }
  mi& operator-=(const mi& m) {
    if ((val -= m.val) < 0) val += M;
    return *this;
  }
  mi& operator*=(const mi& m) {
    val = (ll)val * m.val % M;
    return *this;
  }
  friend mi pow(mi a, ll p) {
    mi ans = 1;
    assert(p >= 0);
    for (; p; p /= 2, a *= a)
      if (p & 1) ans *= a;
    return ans;
  }
  friend mi inv(const mi& a) {
    assert(a != 0);
    return pow(a, M - 2);
  }
  mi& operator/=(const mi& m) { return (*this) *= inv(m); }
  friend mi operator+(mi a, const mi& b) { return a += b; }
  friend mi operator-(mi a, const mi& b) { return a -= b; }
  friend mi operator*(mi a, const mi& b) { return a *= b; }
  friend mi operator/(mi a, const mi& b) { return a /= b; }
};
vi invs, fac, ifac;
void genFac(int SZ) {
  invs.resize(SZ), fac.resize(SZ), ifac.resize(SZ);
  invs[1] = fac[0] = ifac[0] = 1;
  for (int i = (2); i <= (SZ - 1); ++i)
    invs[i] = M - (ll)M / i * invs[M % i] % M;
  for (int i = (1); i <= (SZ - 1); ++i) {
    fac[i] = (ll)fac[i - 1] * i % M;
    ifac[i] = (ll)ifac[i - 1] * invs[i] % M;
  }
}
ll comb(int a, int b) {
  if (a < b || b < 0) return 0;
  return (ll)fac[a] * ifac[b] % M * ifac[a - b] % M;
}
int a[nax][26];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  string second;
  int n;
  re(second, n);
  memset((a), 0, sizeof(a));
  for (int i = (1); i <= (((int)(second).size())); ++i) {
    for (int j = 0; j < (26); ++j) {
      a[i][j] = a[i - 1][j] + (second[i - 1] - 'a' == j ? 1 : 0);
    }
  }
  for (int i = 0; i < (n); ++i) {
    int l, r;
    re(l, r);
    int chk = 0;
    for (int j = 0; j < (26); ++j) {
      if (a[r][j] - a[l - 1][j]) chk++;
    }
    ps(chk > 2 || l == r || second[l - 1] != second[r - 1] ? "YES" : "NO");
  }
}