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;
char s[5555];
int q;
int is[5555][5555] = {};
int main() {
  scanf("%s", s);
  int n = strlen(s);
  for (int len = 1; len < n + 1; len++) {
    for (int i = n - len; i >= 0; i--) {
      int j = i + len - 1;
      if (s[i] == s[j]) {
        if (len <= 3)
          is[i][j] = 1;
        else {
          is[i][j] = is[i + 1][j - 1];
        }
      }
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) is[i][j] += is[i][j - 1];
  for (int i = 1; i < n; i++)
    for (int j = i - 1; j < n; j++) is[i][j] += is[i - 1][j];
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    printf("%d\n", is[r][r] - (l == 0 ? 0 : is[l - 1][r]));
  }
  return 0;
}
