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;
int mod = 1e9 + 7;
int powmod(long long a, int b) {
  long long res = 1;
  if (a >= mod) a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    if (res >= mod) res %= mod;
    a = a * a;
    if (a >= mod) a %= mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int N = 5002;
int h1[N][N], rh1[N][N] = {0};
long long mod1, mod2;
int main() {
  ios_base::sync_with_stdio(false);
  ;
  cin.tie(0);
  mod1 = 1e9 + 6;
  mod2 = 1e9 + 9;
  int i, j, t;
  string x;
  long long a, b, c, d;
  cin >> x;
  for (i = 0; i < x.length(); i++) {
    a = 0;
    b = 0;
    for (j = i; j < x.length(); j++) {
      a = a * 31 + (x[j] - 'a') + 1;
      if (a >= mod1) a %= mod1;
      h1[i][j] = a;
    }
  }
  for (i = x.length() - 1; i >= 0; i--) {
    a = 0;
    b = 0;
    for (j = i; j >= 0; j--) {
      a = a * 31 + (x[j] - 'a' + 1);
      if (a >= mod1) a %= mod1;
      rh1[j][i] = a;
    }
  }
  for (i = 0; i < x.length(); i++) {
    for (j = i; j < x.length(); j++) {
      if (h1[i][j] == rh1[i][j]) {
        h1[i][j] = 1;
      } else
        h1[i][j] = 0;
    }
  }
  for (i = 0; i < x.length(); i++) {
    for (j = 0; j < x.length(); j++) {
      if (i != 0) h1[i][j] += h1[i - 1][j];
    }
  }
  for (i = 0; i < x.length(); i++) {
    for (j = 0; j < x.length(); j++) {
      if (j != 0) h1[i][j] += h1[i][j - 1];
    }
  }
  int q;
  cin >> q;
  int l, r;
  while (q--) {
    cin >> l >> r;
    l--;
    r--;
    if (l != 0)
      cout << h1[r][r] - h1[l - 1][r] - h1[r][l - 1] + h1[l - 1][l - 1] << '\n';
    else
      cout << h1[r][r] << '\n';
  }
  return 0;
}
