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 MM = 5e3 + 5;
string s;
int n, q;
long long dp[MM][MM];
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> s;
  cin >> q;
  n = s.size();
  s = ' ' + s;
  for (int i = 1; i <= n; i++) dp[i][i] = 1;
  for (int len = 2; len <= n; len++) {
    for (int i = 1; i + len - 1 <= n; i++) {
      int j = i + len - 1;
      if (len == 2 && s[i] == s[j])
        dp[i][j] = 1;
      else if (s[i] == s[j])
        dp[i][j] |= dp[i + 1][j - 1];
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      if (j < i) assert(dp[i][j] == 0);
      dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + dp[i][j];
    }
  auto get = [&](int x1, int x2, int y1, int y2) -> long long {
    return dp[x2][y2] - dp[x1 - 1][y2] - dp[x2][y1 - 1] + dp[x1 - 1][y1 - 1];
  };
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << get(l, r, l, r) << "\n";
  }
}
