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;
const long long int infinity = 9e18;
bool dp1[5001][5001] = {};
int dp2[5001][5001] = {};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  cin >> s;
  int n = s.size();
  for (int i = 0; i < n; i++) {
    dp1[i][i] = 1;
    dp2[i][i] = 1;
  }
  for (int len = 2; len <= n; len++) {
    for (int i = 0; i < n - len + 1; i++) {
      int j = i + len - 1;
      if (len == 2) {
        if (s[i] == s[j]) {
          dp1[i][j] = 1;
        } else {
          dp1[i][j] = 0;
        }
        dp2[i][j] = 2 + dp1[i][j];
      } else {
        dp1[i][j] = (dp1[i + 1][j - 1] && s[i] == s[j]);
        dp2[i][j] =
            dp2[i + 1][j] + dp2[i][j - 1] - dp2[i + 1][j - 1] + dp1[i][j];
      }
    }
  }
  long long int q;
  cin >> q;
  long long int a, b;
  for (int i = 0; i < q; i++) {
    cin >> a >> b;
    a--, b--;
    cout << dp2[a][b] << "\n";
  }
  return 0;
}
