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 N = 5000;
int odd[N], even[N];
string s;
void calc() {
  int n = s.length();
  for (int i = 0; i < n; ++i) {
    int l = i, r = i;
    while (l >= 0 && r < n && s[l] == s[r]) --l, ++r;
    ++l;
    --r;
    odd[i] = i - l;
  }
  for (int i = 0; i < n - 1; ++i) {
    int l = i, r = i + 1;
    while (l >= 0 && r < n && s[l] == s[r]) --l, ++r;
    ++l;
    --r;
    even[i] = i - l;
  }
}
int sum[N][N];
bool f[N][N];
void init() {
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < N; ++j) sum[i][j] = 0;
  for (int l = 0; l < s.length(); ++l)
    for (int r = l; r < s.length(); ++r)
      if ((r - l) % 2 == 0) {
        int mid = (l + r) / 2;
        if (mid - l <= odd[mid]) sum[l][r] = 1;
      } else {
        int mid = (l + r) / 2;
        if (mid - l <= even[mid]) sum[l][r] = 1;
      }
  for (int len = 2; len <= s.length(); ++len)
    for (int l = 0; l + len - 1 < s.length(); ++l) {
      int r = l + len - 1;
      sum[l][r] += (sum[l + 1][r] + sum[l][r - 1] - sum[l + 1][r - 1]);
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s;
  calc();
  init();
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    cout << sum[l][r] << "\n";
  }
  return 0;
}
