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 DAY[13] = {-1, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
string s;
int n, q, l, r, pa[5001][5001];
int main() {
  if (false) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; i += 1) {
    pa[i][i] = 1;
    if (i == n - 1) continue;
    if (s[i] == s[i + 1])
      pa[i][i + 1] = 1;
    else
      pa[i][i + 1] = 0;
  }
  for (int len = 2; len < n; len += 1) {
    for (int i = 0; i < n - len; i += 1) {
      pa[i][i + len] = ((s[i] == s[i + len]) && pa[i + 1][i + len - 1]);
    }
  }
  for (int len = 0; len < n; len += 1) {
    for (int i = 0; i < n - len; i += 1) {
      pa[i][i + len] +=
          pa[i + 1][i + len] + pa[i][i + len - 1] - pa[i + 1][i + len - 1];
    }
  }
  cin >> q;
  for (int i = 0; i < q; i += 1) {
    scanf("%d %d", &l, &r);
    l--;
    r--;
    printf("%d\n", pa[l][r]);
  }
  return 0;
}
