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;
bool dp2[5002][5002];
int dp1[5002][5002];
void gen(string s) {
  int n = s.size();
  for (int i = 1; i <= n; i++) {
    dp2[i][i] = true;
    dp1[i][i] = 1;
  }
  for (int i = 1; i < n; i++) {
    dp2[i][i + 1] = (s[i] == s[i - 1]);
    dp1[i][i + 1] = (s[i] == s[i - 1]) + 2;
  }
  for (int l = 3; l <= n; l++)
    for (int i = 0; i <= n - l; i++) {
      dp2[i + 1][i + 1 + l - 1] =
          (s[i] == s[i + l - 1] && dp2[i + 2][i + 1 + l - 1 - 1]);
      dp1[i + 1][i + 1 + l - 1] = (dp2[i + 1][i + 1 + l - 1]) +
                                  dp1[i + 2][i + 1 + l - 1] +
                                  dp1[i + 1][i + l - 1] - dp1[i + 2][i + l - 1];
    }
}
int main() {
  char s[5004];
  scanf("%s", s);
  gen(s);
  int q;
  scanf("%d", &q);
  while (q--) {
    int i, j;
    scanf("%d%d", &i, &j);
    printf("%d\n", dp1[i][j]);
  }
  return 0;
}
