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;
int N, Q, pal[5001][5001], pre[5001][5001];
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  N = s.size();
  s = ' ' + s;
  for (int i = 1; i <= N; i++) {
    pal[i][i] = 1;
  }
  for (int i = 1; i < N; i++) {
    if (s[i] == s[i + 1]) {
      pal[i][i + 1] = 1;
    }
  }
  for (int len = 0; len < N - 2; len++) {
    for (int i = 2; i < N - len; i++) {
      if (pal[i][i + len] == 1 && s[i - 1] == s[i + len + 1]) {
        pal[i - 1][i + len + 1] = 1;
      }
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) {
      pre[i][j] = pal[i][j] + pre[i - 1][j] + pre[i][j - 1] - pre[i - 1][j - 1];
    }
  }
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int a, b;
    cin >> a >> b;
    cout << pre[b][b] - pre[a - 1][b] - pre[b][a - 1] + pre[a - 1][a - 1]
         << "\n";
  }
}
