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 base = 3137;
string s;
int dp[5001][5001];
long long prefix_hash[5001];
long long sufix_hash[5001];
long long pot[5001];
void make_prefix_hash() {
  prefix_hash[0] = int(s[0]);
  pot[0] = 1;
  for (int i = 1; i < s.size(); i++) {
    pot[i] = pot[i - 1] * base;
    prefix_hash[i] = prefix_hash[i - 1] * base + int(s[i]);
  }
}
void make_sufix_hash() {
  sufix_hash[s.size() - 1] = int(s[s.size() - 1]);
  for (int i = s.size() - 2; i >= 0; i--) {
    sufix_hash[i] = sufix_hash[i + 1] * base + int(s[i]);
  }
}
int palin(int l, int r) {
  long long dhash, lhash;
  if (!l)
    lhash = prefix_hash[r];
  else
    lhash = prefix_hash[r] - pot[r - l + 1] * prefix_hash[l - 1];
  if (r == s.size() - 1)
    dhash = sufix_hash[l];
  else
    dhash = sufix_hash[l] - pot[r - l + 1] * sufix_hash[r + 1];
  if (lhash == dhash)
    return 1;
  else
    return 0;
}
void sol() {
  make_prefix_hash();
  make_sufix_hash();
  for (int i = 0; i < s.size(); i++) dp[i][i] = 1;
  for (int dulj = 1; dulj < s.size(); dulj++)
    for (int i = 0; i + dulj < s.size(); i++) {
      dp[i][i + dulj] =
          dp[i + 1][i + dulj] + dp[i][i + dulj - 1] + palin(i, i + dulj);
      if (dulj > 1) dp[i][i + dulj] -= dp[i + 1][i + dulj - 1];
    }
}
int main() {
  cin >> s;
  int num;
  cin >> num;
  sol();
  while (num--) {
    int l, r;
    scanf("%d %d", &l, &r);
    l--;
    r--;
    printf("%d\n", dp[l][r]);
  }
  return 0;
}
