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;
const int mx = 5000 + 1;
bool isPalindrome[mx][mx];
int dp[mx][mx];
char s[21000];
int main() {
  scanf("%s", s);
  int q;
  scanf("%d", &q);
  int l = strlen(s);
  for (int len = 0; len < l; ++len) {
    for (int i = 0; i < l; ++i) {
      if (i + len >= l) break;
      if (len == 0) isPalindrome[i][len] = 1;
      if (len == 1)
        isPalindrome[i][len] = (s[i] == s[i + len]) ? true : false;
      else if (s[i] == s[i + len] && isPalindrome[i + 1][len - 2])
        isPalindrome[i][len] = 1;
    }
  }
  for (int len = 0; len < l; ++len) {
    for (int i = 0; i < l; ++i) {
      if (i + len >= l) break;
      if (len == 0)
        dp[i][len] = 1;
      else if (len == 1)
        dp[i][len] = dp[i][0] + dp[i + 1][0] + isPalindrome[i][len];
      else
        dp[i][len] = dp[i][len - 1] + dp[i + 1][len - 1] +
                     isPalindrome[i][len] - dp[i + 1][len - 2];
    }
  }
  for (int i = 0; i < q; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    printf("%d\n", dp[a][b - a]);
  }
}
