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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long bin_pow(long long a, long long n) {
  if (n == 0) return 1;
  if (n & 1) return a * bin_pow(a, n - 1);
  long long b = bin_pow(a, n / 2);
  return b * b;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  string s;
  cin >> s;
  int n = s.size();
  vector<vector<long long>> dp(s.size(), vector<long long>(s.size()));
  vector<vector<bool>> isp(s.size(), vector<bool>(s.size()));
  for (int i = 0; i < s.size(); i++) {
    int j = i;
    dp[i][j] = 1;
    isp[i][j] = 1;
  }
  for (int i = 0; i + 1 < s.size(); i++) {
    int j = i + 1;
    isp[i][j] = (s[i] == s[j]);
    dp[i][j] = 2 + isp[i][j];
  }
  for (int len = 2; len < s.size(); len++) {
    for (int i = 0; i + len < s.size(); i++) {
      int j = i + len;
      isp[i][j] = (s[i] == s[j]) & isp[i + 1][j - 1];
      dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1] + isp[i][j];
    }
  }
  int t;
  cin >> t;
  while (t--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << dp[l][r] << '\n';
  }
  return 0;
}
