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>
using namespace std;
template <typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
  for (auto it : v) s << it << " ";
  return s;
}
template <typename T, typename T2>
ostream &operator<<(ostream &s, const pair<T, T2> &p) {
  return s << "(" << p.fi << "," << p.se << ")";
}
struct debugger {
  template <typename T>
  debugger &operator,(const T &v) {
    cerr << v << ", ";
    return *this;
  }
} dbg;
int dp[5000][5000];
int is_pal[5000][5000];
string str;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> str;
  for (int i = 0, i_ = (str.size()); i < i_; i++) {
    dp[i][i] = 1;
    is_pal[i][i] = 1;
    is_pal[i + 1][i] = 1;
  }
  for (int i = (2), i_ = (str.size() + 1); i < i_; i++)
    for (int j = 0, j_ = (str.size() - i + 1); j < j_; j++) {
      is_pal[j][j + i - 1] =
          is_pal[j + 1][j + i - 2] and str[j] == str[j + i - 1];
      dp[j][j + i - 1] = dp[j + 1][j + i - 1] + dp[j][j + i - 2] -
                         dp[j + 1][j + i - 2] + is_pal[j][j + i - 1];
    }
  int n;
  cin >> n;
  for (int i = 0, i_ = (n); i < i_; i++) {
    int l, r;
    cin >> l >> r;
    cout << dp[l - 1][r - 1] << endl;
  }
  return 0;
}
