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 dp[5000][5000] = {0};
bool p[5000][5000] = {0};
string s;
int q, l, r, n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    cin >> s;
    cin >> q;
    n = s.size();
    memset(dp, 0, sizeof(dp));
    memset(p, false, sizeof(p));
    for (long long int i = 0; i < n; i++) p[i][i] = true;
    for (long long int i = 0; i < n - 1; i++)
      if (s[i] == s[i + 1]) p[i][i + 1] = true, dp[i][i + 1] = 1;
    for (long long int g = 2; g < n; g++)
      for (long long int i = 0; i < n - g; i++) {
        int j = i + g;
        if (s[i] == s[j] && p[i + 1][j - 1]) p[i][j] = true;
        if (p[i][j] == true)
          dp[i][j] = dp[i][j - 1] + 1 + dp[i + 1][j] - dp[i + 1][j - 1];
        else
          dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1];
      }
    for (long long int i = 0; i < q; i++) {
      cin >> l >> r;
      cout << dp[l - 1][r - 1] + r - l + 1 << '\n';
    }
  }
  return 0;
}
