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 q;
int ok[5010][5010], d[5010][5010];
bool check(int l, int r) {
  if (l - r >= 1) return 1;
  if (ok[l][r] != -1) return ok[l][r];
  if (s[l] == s[r]) {
    if (check(l + 1, r - 1)) return ok[l][r] = 1;
  } else
    return ok[l][r] = 0;
}
int calc(int l, int r) {
  if (l > r) return 0;
  if (d[l][r] != -1) return d[l][r];
  return d[l][r] =
             check(l, r) + calc(l, r - 1) + calc(l + 1, r) - calc(l + 1, r - 1);
}
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  scanf("%d", &q);
  for (int i = 1; i <= n; i++) {
    ok[i][i] = 1;
    if (i + 1 <= n && s[i] == s[i + 1]) ok[i][i + 1] = 1;
  }
  for (int len = 3; len <= n; len++) {
    for (int l = 1; l + len - 1 <= n; l++) {
      int r = l + len - 1;
      if (ok[l + 1][r - 1] && s[l] == s[r]) ok[l][r] = 1;
    }
  }
  for (int len = 1; len <= n; len++) {
    for (int l = 1; l + len - 1 <= n; l++) {
      int r = l + len - 1;
      d[l][r] = ok[l][r] + d[l][r - 1] + d[l + 1][r] - d[l + 1][r - 1];
    }
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", d[l][r]);
  }
  return 0;
}
