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>
clock_t t_;
using namespace std;
template <class T>
void print(T s) {
  cout << s << "\n";
}
template <class T>
void emax(T &a, T b) {
  a = (a < b ? b : a);
};
template <class T>
void emin(T &a, T b) {
  a = (a > b ? b : a);
}
template <class T>
void printC(T s) {
  for (auto &x : s) {
    cout << x << " ";
  }
  cout << "\n";
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
string s;
int n;
int dp[5010][5010], pal[5010][5010];
int main(int argc, const char *argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0), cin.tie(0), cout.tie(0),
      cout.precision(15);
  t_ = clock();
  cin >> s;
  n = (int)s.length();
  int q;
  cin >> q;
  for (int i = int(0); i < int(n); i++) {
    dp[i][i] = 1LL;
    pal[i][i] = 1;
  }
  int l, r;
  for (int i = int(0); i < int(n); i++) {
    l = i, r = i;
    pal[l][r] = 1;
    while (l - 1 >= 0 && r + 1 < n && s[l - 1] == s[r + 1]) {
      l--, r++;
      pal[l][r] = 1;
    }
  }
  for (int i = int(0); i < int(n - 1); i++) {
    if (s[i] == s[i + 1]) {
      l = i, r = i + 1;
      pal[l][r] = 1;
      while (l - 1 >= 0 && r + 1 < n && s[l - 1] == s[r + 1]) {
        l--, r++;
        pal[l][r] = 1;
      }
    }
  }
  for (int i = int(2); i < int(n + 1); i++) {
    for (int j = int(0); j < int(n); j++) {
      if (j + i - 1 < n) {
        dp[j][j + i - 1] = dp[j][j + i - 2] + dp[j + 1][j + i - 1] +
                           pal[j][i + j - 1] - (dp[j + 1][j + i - 2]);
      }
    }
  }
  for (int i = int(0); i < int(q); i++) {
    cin >> l >> r;
    l--, r--;
    cout << dp[l][r] << "\n";
  }
  t_ = clock() - t_;
  cerr << setprecision(9) << fixed << ((float)t_ / CLOCKS_PER_SEC) << endl;
  return 0;
}
