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>
char s[5005];
bool f[5005][5005];
int dp[5005][5005];
int main() {
  int len, q, left, right;
  scanf("%s", s);
  len = strlen(s);
  for (int l = 0; l < len; l++) {
    for (int i = 0; i + l < len; i++) {
      int j = i + l;
      if (l == 0) {
        f[i][j] = true;
        dp[i][j] = 1;
        continue;
      }
      if (l == 1) {
        if (s[i] == s[j]) {
          f[i][j] = true;
          dp[i][j] = 3;
        } else {
          dp[i][j] = 2;
        }
        continue;
      }
      if (f[i + 1][j - 1] && s[i] == s[j]) {
        f[i][j] = true;
      }
      dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1];
      if (f[i][j]) {
        dp[i][j]++;
      }
    }
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &left, &right);
    printf("%d\n", dp[left - 1][right - 1]);
  }
  return 0;
}
