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>
int f[5010][5010];
int N;
char s[5010];
int l, r;
int len;
int is[5010][5010];
int ish(int i, int j) {
  if (i >= j) return is[i][j] = 1;
  if (is[i][j] != -1) return is[i][j];
  return is[i][j] = (s[i] == s[j]) && ish(i + 1, j - 1);
}
void dp() {
  for (int i = 1; i <= len; i++) f[i][i] = 1;
  for (int i = 1; i <= len - 1; i++) f[i][i + 1] = 2 + ish(i, i + 1);
  for (int k = 3; k <= len; k++)
    for (int i = 1; i + k - 1 <= len; i++) {
      f[i][i + k - 1] = f[i + 1][i + k - 1] + f[i][i + k - 2] -
                        f[i + 1][i + k - 2] + ish(i, i + k - 1);
    }
}
int main() {
  scanf("%s", s + 1);
  scanf("%d", &N);
  len = 1;
  while (s[len] != '\0') {
    len++;
  }
  memset(f, 0, sizeof(f));
  memset(is, -1, sizeof(is));
  len--;
  dp();
  for (int i = 1; i <= N; i++) {
    scanf("%d%d", &l, &r);
    printf("%d\n", f[l][r]);
  }
}
