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 g[5007][5007];
int d[2][5007];
void Man(string &S, bool T) {
  int n = S.size();
  for (int i = 0, l = 0, r = -1; i < n; i++) {
    int k = (i > r) ? T : min(d[T][l + r - i + !T], r - i + 1);
    while (0 <= i - k - !T && i + k < n && S[i - k - !T] == S[i + k]) k++;
    d[T][i] = k--;
    if (i + k > r) {
      l = i - k - !T;
      r = i + k;
    }
  }
}
bool isPal(int l, int r) {
  return (d[((r - l + 1) & 1)][((r + l - 1) >> 1)] >= ((r - l + 2) >> 1));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  Man(s, 0);
  Man(s, 1);
  int n = s.size();
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
      if (isPal(i, j))
        g[1][j]++, g[1][n + 1]--, g[i + 1][j]--, g[i + 1][n + 1]++;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      g[i][j] += g[i - 1][j] + g[i][j - 1] - g[i - 1][j - 1];
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << g[l][r] << '\n';
  }
}
