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 long long mod = 1e9 + 7;
long long add(long long a, long long b) { return (a % mod + b % mod) % mod; }
long long mult(long long a, long long b) { return (a % mod * b % mod) % mod; }
long long sub(long long a, long long b) {
  return (a % mod - b % mod + mod) % mod;
}
long long dp[5000][5000];
bool palin[5000][5000];
void solve() {
  string s;
  cin >> s;
  long long n = s.length();
  for (long long i = 0; i < n; i++) palin[i][i] = true;
  for (long long i = 0; i + 1 < n; i++)
    if (s[i] == s[i + 1]) palin[i][i + 1] = true;
  for (long long i = 3; i <= n; i++)
    for (long long j = 0; j + i - 1 < n; j++)
      palin[j][j + i - 1] = (palin[j + 1][j + i - 2] && s[j] == s[j + i - 1]);
  for (long long i = 0; i < n; i++) dp[i][i] = 1;
  for (long long i = 0; i + 1 < n; i++) dp[i][i + 1] = 2 + (s[i] == s[i + 1]);
  for (long long i = 3; i <= n; i++)
    for (long long j = 0; j + i - 1 < n; j++)
      dp[j][j + i - 1] = palin[j][j + i - 1] + dp[j][j + i - 2] +
                         dp[j + 1][j + i - 1] - dp[j + 1][j + i - 2];
  long long q;
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    l--, r--;
    cout << dp[l][r] << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long tc = 1;
  for (long long i = 1; i <= tc; i++) {
    solve();
  }
}
