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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
char dp[5100][5100];
short w[5100][5100];
int res[5100][5100];
int main() {
  string s;
  cin >> s;
  int len = s.length();
  for (int i = 0; i < len; i++) {
    for (int j = 0; i - j >= 0 && i + j < len; j++)
      if (s[i - j] != s[i + j])
        break;
      else
        dp[j * 2 + 1][i + j] = 1;
    for (int j = 0; i - j >= 0 && i + j + 1 < len; j++)
      if (s[i - j] != s[i + j + 1])
        break;
      else
        dp[j * 2 + 2][i + j + 1] = 1;
  }
  for (int i = 0; i < len; i++)
    for (int j = 1; j <= len; j++) w[i][j] = w[i][j - 1] + dp[j][i];
  for (int i = 0; i < len; i++) {
    res[i][i] = w[i][1];
    for (int j = i + 1; j < len; j++)
      res[i][j] = res[i][j - 1] + w[j][j - i + 1];
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    cout << res[a][b] << endl;
  }
  return 0;
}
