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;
string s;
int n, a, b;
int d[5050][5050], val[5050][5050];
void init() {
  int left, right;
  int x = s.size();
  for (int i = 0; i < x; i++) d[i][i] = 1;
  for (int len = 2; len <= x; len++)
    for (int i = 0; i + len - 1 < x; i++) {
      left = i, right = i + len - 1;
      if (len == 2)
        d[left][right] = s[left] == s[right];
      else
        d[left][right] = d[left + 1][right - 1] and (s[left] == s[right]);
    }
  for (int i = 0; i < 5030; i++) {
    val[i][i] = 1;
    val[i][i + 1] = 2 + d[i][i + 1];
  }
  for (int len = 3; len <= x; len++)
    for (int i = 0; i + len - 1 < x; i++)
      val[i][i + len - 1] = d[i][i + len - 1] + val[i][i + len - 2] +
                            val[i + 1][i + len - 1] - val[i + 1][i + len - 2];
}
int main() {
  cin >> s;
  init();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    cout << val[a - 1][b - 1] << "\n";
  }
  return 0;
}
