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;
mt19937_64 random_num(
    chrono::high_resolution_clock::now().time_since_epoch().count());
const long long MOD = 1000000007;
const int base = 26;
string s, rs;
int n, q;
bool palin[5001][5001];
long long mu[5001];
long long hashS[5001], rhashS[5001];
long long f[5001][5001];
long long getHash(int i, int j, long long *H) {
  return ((H[j] - (H[i - 1] * mu[j - i + 1]) % MOD + MOD)) % MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  rs = s;
  reverse(rs.begin(), rs.end());
  n = s.size();
  s = ' ' + s;
  rs = ' ' + rs;
  mu[0] = 1;
  for (int i = 1; i <= n; i++) mu[i] = (mu[i - 1] * 26) % MOD;
  for (int i = 1; i <= n; i++) {
    hashS[i] = (hashS[i - 1] * 26 + (s[i] - 'a' + 1)) % MOD;
    rhashS[i] = (rhashS[i - 1] * 26 + (rs[i] - 'a' + 1)) % MOD;
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
      palin[i][j] =
          (getHash(i, j, hashS) == getHash(n - j + 1, n - i + 1, rhashS));
  for (int i = 1; i <= n; i++) {
    f[i][i] = 1;
    if (i > 1) f[i - 1][i] = palin[i - 1][i] + 2;
    for (int j = i - 2; j >= 1; j--)
      f[j][i] = (f[j][i - 1] + f[j + 1][i] - f[j + 1][i - 1]) + palin[j][i];
  }
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << f[l][r] << "\n";
  }
}
