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 MAXN = 5010;
const long long base = 727, MOD = 1e9 + 17;
vector<long long> d, h, hr;
int n, q;
string s;
vector<vector<long long> > dp;
void make_hash(string &s) {
  d.resize(MAXN, 1);
  h.resize(MAXN, 0);
  hr.resize(MAXN, 0);
  for (int i = 1; i < MAXN; i++) d[i] = (d[i - 1] * base) % MOD;
  for (int i = 0; i < n; i++) h[i + 1] = (h[i] * base + s[i]) % MOD;
  for (int i = n - 1; i >= 0; i--) hr[i] = (hr[i + 1] * base + s[i]) % MOD;
}
long long get_h(int l, int r) {
  return (h[r] - h[l - 1] * d[r - l + 1] % MOD + MOD) % MOD;
}
long long get_hr(int l, int r) {
  return (hr[l - 1] - hr[r] * d[r - l + 1] % MOD + MOD) % MOD;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  cin >> s;
  n = (int)s.size();
  make_hash(s);
  dp.resize(MAXN, vector<long long>(MAXN, 0));
  for (int len = 1; len <= n; len++) {
    for (int i = 1; i <= n - len + 1; i++) {
      dp[i][i + len - 1] =
          dp[i][i + len - 2] + dp[i + 1][i + len - 1] - dp[i + 1][i + len - 2];
      if (get_h(i, i + len - 1) == get_hr(i, i + len - 1)) dp[i][i + len - 1]++;
    }
  }
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << dp[l][r] << "\n";
  }
  return 0;
}
