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 setIO(string str, bool dbg) {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  if (!dbg) {
    freopen((str + ".in").c_str(), "r", stdin);
    freopen((str + ".out").c_str(), "w", stdout);
  }
}
const int MAX = 5e3 + 5;
const int LEN = 2e3 + 5;
const int LVL = 2e1 + 0;
const long long int MOD = 1e9 + 7;
const long long int INF = 2e9;
int xyz = 1;
int n, q;
int pld[MAX][MAX];
int dps[MAX][MAX];
void fnc() {
  string str;
  cin >> str;
  n = str.length();
  for (int i = 0; i < n; i++) pld[i][i] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 0; j + i < n; j++) {
      int l = j;
      int r = i + j;
      pld[l][r] = str[l] == str[r] ? pld[l + 1][r - 1] || r - l == 1 : 0;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j + i < n; j++) {
      int l = j;
      int r = i + j;
      dps[l][r] = pld[l][r] + dps[l + 1][r] + dps[l][r - 1] - dps[l + 1][r - 1];
    }
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l;
    int r;
    cin >> l >> r;
    cout << dps[l - 1][r - 1] << '\n';
  }
}
int main() {
  setIO("", 1);
  while (xyz--) fnc();
  return 0;
}
