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_base::sync_with_stdio(0);
  cin.tie(0);
  string s;
  int q, i, j, l, r, k;
  cin >> s >> q;
  int dp[s.length() + 2][s.length() + 1];
  bool isPal[s.length() + 2][s.length() + 1];
  memset(dp, 0, sizeof(dp));
  memset(isPal, true, sizeof(isPal));
  for (k = 1; k <= s.length(); k++) {
    i = 1;
    j = k;
    while (j <= s.length()) {
      if (s[i - 1] == s[j - 1] && isPal[i + 1][j - 1]) {
        isPal[i][j] = true;
      } else
        isPal[i][j] = false;
      if (isPal[i][j]) {
        dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1] + 1;
      } else
        dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1];
      i++;
      j++;
    }
  }
  while (q--) {
    cin >> l >> r;
    cout << dp[l][r] << "\n";
  }
}
