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 palin[5009][5009];
int num[5009][5009];
string inp;
bool isPalin(int i, int j) {
  if (i >= j) return true;
  if (palin[i][j] != -1) return (palin[i][j] == 1);
  if (isPalin(i + 1, j - 1) && inp[i] == inp[j]) {
    palin[i][j] = 1;
    return true;
  } else {
    palin[i][j] = 0;
    return false;
  }
}
int getPallin(int s, int e) {
  if (s > e) return 0;
  if (s == e) return 1;
  if (num[s][e] != -1) return num[s][e];
  int ans = 0;
  if (isPalin(s, e)) ans = 1;
  ans += getPallin(s + 1, e);
  ans += getPallin(s, e - 1);
  ans -= getPallin(s + 1, e - 1);
  return num[s][e] = ans;
}
int main() {
  cin >> inp;
  memset(palin, -1, sizeof(palin));
  memset(num, -1, sizeof(num));
  int q;
  int s, e;
  cin >> q;
  while (q--) {
    cin >> s >> e;
    s--;
    e--;
    cout << getPallin(s, e) << endl;
  }
  return 0;
}
