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 = 5e3 + 1;
string s;
int n;
int a[N];
int c[N][N], d[N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  n = s.length();
  for (int i = 1; i <= n; i++) a[i] = s[i - 1];
  for (int i = 1; i <= n; i++) {
    c[i][i] = d[i][i] = 1;
    if (i < n && a[i] == a[i + 1]) {
      d[i][i + 1] = 3;
      c[i][i + 1] = 1;
    } else
      d[i][i + 1] = 2;
  }
  for (int len = 3; len <= n; len++) {
    for (int i = 1; i <= n - len + 1; i++) {
      int j = i + len - 1;
      d[i][j] = d[i + 1][j] + d[i][j - 1] - d[i + 1][j - 1];
      if (a[i] == a[j] && c[i + 1][j - 1]) {
        d[i][j]++;
        c[i][j] = 1;
      }
    }
  }
  int m, x, y;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    cout << d[x][y] << "\n";
  }
}
