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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  int length = s.length();
  vector<vector<bool> > P(length);
  vector<vector<int> > nr_of_P(length);
  for (int i = 0; i < length; i++) {
    if (i == 0) {
      vector<bool> p(length - i, true);
      vector<int> nr(length - i, 1);
      P[i] = p;
      nr_of_P[i] = nr;
    } else if (i == 1) {
      vector<bool> p(length - i);
      vector<int> nr(length - i);
      for (int j = 0; j < length - i; j++) {
        if (s[j] == s[j + 1]) {
          p[j] = true;
        }
        nr[j] = nr_of_P[i - 1][j] + nr_of_P[i - 1][j + 1] + (int)(p[j] ? 1 : 0);
      }
      P[i] = p;
      nr_of_P[i] = nr;
    } else {
      vector<bool> p(length - i);
      vector<int> nr(length - i);
      for (int j = 0; j < length - i; j++) {
        p[j] = false;
        if (P[i - 2][j + 1] == 1 && s[j] == s[j + i]) p[j] = true;
        nr[j] = nr_of_P[i - 1][j] + nr_of_P[i - 1][j + 1] -
                nr_of_P[i - 2][j + 1] + (int)(p[j] ? 1 : 0);
      }
      P[i] = p;
      nr_of_P[i] = nr;
    }
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    cout << nr_of_P[r - l][l - 1] << "\n";
  }
  return 0;
}
