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;
char s[5010];
int mat[5010][5010];
int dp[5010][5010];
int n;
void findAllPalindromes() {
  for (int i = 0; i < n; i++) {
    for (int j = 0; i - j >= 0 && i + j < n; j++) {
      if (s[i - j] == s[i + j])
        mat[i - j][i + j] = 1;
      else
        break;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; i - j >= 0 && i + j + 1 < n; j++) {
      if (s[i - j] == s[i + j + 1])
        mat[i - j][i + j + 1] = 1;
      else
        break;
    }
  }
}
void filldp() {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      dp[i][j] =
          dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1] + mat[i - 1][j - 1];
    }
}
inline int sum(int l, int r) {
  int ans = dp[n][r];
  if (l > 0) ans -= dp[l - 1][r];
  return ans;
}
int main() {
  scanf(" %s", s);
  n = strlen(s);
  findAllPalindromes();
  filldp();
  int q;
  scanf(" %d", &q);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%d\n", sum(l, r));
  }
  return 0;
}
