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;
void FastIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(15);
}
int dp[5010][5010] = {0}, dp1[5010][5010] = {0};
int ispalin(string &s, int i, int j) {
  if (j < i) return 0;
  if (dp[i][j] != -1) return dp[i][j];
  if (i == j) return 1;
  if (j == i + 1) {
    if (s[i] == s[j]) return 1;
    return 0;
  }
  if (s[i] != s[j]) return 0;
  return s[i] == s[j] ? (dp[i + 1][j - 1] = ispalin(s, i + 1, j - 1)) : 0;
}
int main() {
  FastIO();
  string s;
  cin >> s;
  int l = s.length(), i, j, r, q, sum = 0;
  vector<int> cum;
  cum.resize(l);
  for (i = 0; i < l; i++)
    for (j = 0; j < l; j++) dp[i][j] = -1;
  for (i = 0; i < l; i++)
    for (j = i; j < l; j++) dp[i][j] = ispalin(s, i, j);
  for (i = 0; i < l; i++) {
    for (j = 0; j < l; j++) {
      if (dp[i][j] == -1) dp[i][j] = 0;
    }
  }
  for (i = 0; i < l; i++) {
    sum = 0;
    for (j = 0; j < l; j++) {
      sum = sum + dp[i][j];
      if (i == 0)
        dp1[i][j] = sum;
      else
        dp1[i][j] = dp1[i - 1][j] + sum;
    }
  }
  cin >> q;
  while (q--) {
    cin >> l >> r;
    l--;
    r--;
    if (l == 0)
      cout << dp1[r][r] << "\n";
    else
      cout << dp1[r][r] - dp1[l - 1][r] - dp1[r][l - 1] + dp1[l - 1][l - 1]
           << "\n";
  }
  return 0;
}
