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 MX = 100000;
const long long MOD = 1000000007;
long long fast_exp(long long base, long long exp, long long mod) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % mod;
    base = (base * base) % mod;
    exp /= 2;
  }
  return res;
}
int dp[5010][5010];
int pal[5010][5010];
char s[5010];
int q, l, r, n;
int main() {
  ios_base::sync_with_stdio(false);
  ;
  cin.tie(NULL);
  cin >> s;
  n = strlen(s);
  for (int i = 0, _i = (n); i < _i; ++i) {
    dp[i][i] = pal[i][i] = 1;
    pal[i + 1][i] = 1;
  }
  for (int len = 2; len <= n; len++) {
    for (int start = 0; start <= n - len; start++) {
      pal[start][start + len - 1] =
          pal[start + 1][start + len - 2] & (s[start] == s[start + len - 1]);
      dp[start][start + len - 1] =
          dp[start][start + len - 2] + dp[start + 1][start + len - 1] -
          dp[start + 1][start + len - 2] + pal[start][start + len - 1];
    }
  }
  cin >> q;
  while (q--) {
    cin >> l >> r;
    --l, --r;
    cout << dp[l][r] << "\n";
  }
  return 0;
}
