You've got a string s = s1s2... s|s| of length |s|, consisting of lowercase English letters. There also are q queries, each query is described by two integers li, ri (1 ≤ li ≤ ri ≤ |s|). The answer to the query is the number of substrings of string s[li... ri], which are palindromes.

String s[l... r] = slsl + 1... sr (1 ≤ l ≤ r ≤ |s|) is a substring of string s = s1s2... s|s|.

String t is called a palindrome, if it reads the same from left to right and from right to left. Formally, if t = t1t2... t|t| = t|t|t|t| - 1... t1.

Input

The first line contains string s (1 ≤ |s| ≤ 5000). The second line contains a single integer q (1 ≤ q ≤ 106) — the number of queries. Next q lines contain the queries. The i-th of these lines contains two space-separated integers li, ri (1 ≤ li ≤ ri ≤ |s|) — the description of the i-th query.

It is guaranteed that the given string consists only of lowercase English letters.

Output

Print q integers — the answers to the queries. Print the answers in the order, in which the queries are given in the input. Separate the printed numbers by whitespaces.

Examples

Input

caaaba
5
1 1
1 4
2 3
4 6
4 5


Output

1
7
3
4
2

Note

Consider the fourth query in the first test case. String s[4... 6] = «aba». Its palindrome substrings are: «a», «b», «a», «aba».
#include <bits/stdc++.h>
#pragma optimize("", off)
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/STACK:36777216")
const long long inf = 0x3f3f3f3f3f3f3f3LL;
const long long mod = 1e9 + 7;
using namespace std;
template <class T>
void smin(T& a, T val) {
  if (a > val) a = val;
}
template <class T>
void smax(T& a, T val) {
  if (a < val) a = val;
}
template <typename T>
inline std::ostream& operator<<(std::ostream& os, const std::vector<T>& v) {
  bool first = true;
  os << "[";
  for (auto i : v) {
    if (!first) os << ", ";
    os << i;
    first = false;
  }
  return os << "]";
}
const long long N = 5e3 + 10;
long long ans[N][N];
struct PolynomialHashing {
  long long N;
  string s;
  char offset = 0;
  long long prime;
  long long *fHash, *rHash, *pk;
  PolynomialHashing(string str, long long pri = 257) : s(str), prime(pri) {
    N = s.size();
    fHash = new long long[N], rHash = new long long[N], pk = new long long[N];
    fHash[0] = s[0] - offset + 1;
    pk[0] = 1;
    rHash[N - 1] = s[N - 1] - offset + 1;
    for (long long i = 1; i < N; i++) {
      fHash[i] = ((fHash[i - 1] * prime) % mod + s[i] - offset + 1) % mod;
      pk[i] = (pk[i - 1] * prime) % mod;
      rHash[N - 1 - i] =
          ((rHash[N - i] * prime) % mod + s[N - i - 1] - offset + 1) % mod;
    }
  }
  long long getFrontHash(long long l, long long r) {
    if (l == 0) return fHash[r];
    if (l > r) return 0;
    return (fHash[r] - (fHash[l - 1] * pk[r - l + 1]) % mod + mod) % mod;
  }
  long long getReverseHash(long long l, long long r) {
    if (r == N - 1) return rHash[l];
    if (l > r) return 0;
    return (rHash[l] - (rHash[r + 1] * pk[r - l + 1]) % mod + mod) % mod;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  long long n = (long long)(s.size());
  PolynomialHashing p(s);
  for (long long i = 0; i < n; ++i) {
    for (long long j = i; j < n; ++j) {
      if (j - 1 >= 0) {
        ans[i][j] += ans[i][j - 1];
      }
      long long tt = p.getFrontHash(i, j);
      long long tt2 = p.getReverseHash(i, j);
      ans[i][j] += (tt == tt2);
    }
  }
  for (long long j = n - 1; j >= 0; --j) {
    for (long long i = j - 1; i >= 0; --i) {
      ans[i][j] += ans[i + 1][j];
    }
  }
  long long q, l, r;
  cin >> q;
  while (q--) {
    cin >> l >> r;
    --l, --r;
    cout << ans[l][r] << '\n';
  }
  return 0;
}
