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 main() {
  string s;
  cin >> s;
  int n = s.size();
  vector<vector<int>> P(n + 1, vector<int>(n + 1, 0));
  for (int len = 1; len <= n; len++) {
    for (int start = 1; start <= n - len + 1; start++) {
      if (len == 1 || len == 2) {
        P[start][start + len - 1] = (s[start - 1] == s[start + len - 2]);
      } else {
        P[start][start + len - 1] = (s[start - 1] == s[start + len - 2] &&
                                     P[start + 1][start + len - 2]);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      P[i][j] = P[i][j] + P[i][j - 1] + P[i - 1][j] - P[i - 1][j - 1];
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < (q); i++) {
    int lo, hi;
    scanf("%d %d", &lo, &hi);
    cout << (P[hi][hi] - P[lo - 1][hi] - P[hi][lo - 1] + P[lo - 1][lo - 1])
         << endl;
  }
}
