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 dp[5001][5001];
char s[5001];
int main() {
  scanf("%s", s);
  int n;
  scanf("%d", &n);
  for (int i = 0; *(s + i); i++) {
    dp[i][i] = 1;
    for (int j1 = i - 1, j2 = i + 1; j1 >= 0 && *(s + j2); j1--, j2++)
      if (s[j1] == s[j2])
        dp[j1][j2] = 1;
      else
        break;
    if (s[i] == s[i + 1]) {
      dp[i][i + 1] = 3;
      for (int j1 = i - 1, j2 = i + 2; j1 >= 0 && *(s + j2) && *(s + j2 - 1);
           j1--, j2++)
        if (s[j1] == s[j2])
          dp[j1][j2] = 1;
        else
          break;
    } else
      dp[i][i + 1] = 2;
  }
  for (int k = 2; *(s + k); k++)
    for (int i = 0; *(s + i + k); i++)
      dp[i][i + k] +=
          dp[i][i + k - 1] + dp[i + 1][i + k] - dp[i + 1][i + k - 1];
  while (n--) {
    int i, j;
    scanf("%d%d", &i, &j);
    printf("%d\n", dp[i - 1][j - 1]);
  }
  return 0;
}
