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 Q, hd[5005], ht[5005], prm[5005], n = 1, dyn[5005][5005];
char str[5005];
int pal(int b, int e) {
  return (hd[e] - hd[b - 1]) * prm[n - e] == (ht[b] - ht[e + 1]) * prm[b - 1];
}
int fun(int bgn, int end) {
  if (dyn[bgn][end] != -1) return dyn[bgn][end];
  if (end == bgn) return dyn[bgn][end] = 1;
  if (end - 1 == bgn) return dyn[bgn][end] = pal(bgn, end) + 2;
  return dyn[bgn][end] = fun(bgn + 1, end) + fun(bgn, end - 1) -
                         fun(bgn + 1, end - 1) + pal(bgn, end);
}
int main() {
  scanf(" %s", str + 1);
  n = strlen(str + 1);
  prm[0] = 1;
  memset(dyn, -1, sizeof(dyn));
  for (int i = 1; i <= n; i++) prm[i] = prm[i - 1] * 37;
  for (int i = 1; i <= n; i++)
    hd[i] = hd[i - 1] + prm[i - 1] * (str[i] - 'a' + 1);
  for (int i = n; i > 0; i--)
    ht[i] = ht[i + 1] + prm[n - i] * (str[i] - 'a' + 1);
  scanf("%d", &Q);
  for (int i = 1, a, b; i <= Q; i++) {
    scanf("%d %d", &a, &b);
    printf("%d\n", fun(a, b));
  }
}
