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 p[5002][5002];
int ans[5002][5002];
int main() {
  string s;
  cin >> s;
  for (int i = 0; i < 5002; i++)
    for (int j = 0; j < 5002; j++) p[i][j] = 0;
  for (int i = 0; i < s.size(); i++) {
    for (int j = 0; (i - j) >= 0 && ((i + j) < s.size()); j++) {
      if (s[i - j] == s[i + j])
        p[i - j][i + j]++;
      else
        break;
    }
    for (int j = 0; (i - j) >= 0 && ((i + j + 1) < s.size()); j++) {
      if (s[i - j] == s[i + j + 1])
        p[i - j][i + j + 1]++;
      else
        break;
    }
  }
  for (int i = 0; i < 5002; i++) {
    for (int j = i; j < 5002; j++) {
      p[i][j] = p[i][j];
      if (i != 0) p[i][j] += p[i - 1][j];
    }
  }
  for (int i = 0; i < 5002; i++) {
    for (int j = i; j < 5002; j++) {
      ans[i][j] = p[j][j];
      if (j > 0) ans[i][j] += ans[i][j - 1];
      if (i > 0) ans[i][j] -= p[i - 1][j];
    }
  }
  int n;
  cin >> n;
  int l, r;
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &l, &r);
    printf("%d\n", ans[l - 1][r - 1]);
  }
  return 0;
}
