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 main() {
  ios::sync_with_stdio(0);
  string s;
  cin >> s;
  vector<vector<bool> > dp(s.size());
  vector<vector<int> > num(s.size());
  for (__typeof(s.size()) i = 0; i < (s.size()); ++i) {
    dp[i].resize(s.size());
    dp[i][i] = true;
    num[i].resize(s.size());
    num[i][i] = 1;
  }
  for (__typeof(s.size() - 1) i = 0; i < (s.size() - 1); ++i) {
    dp[i][i + 1] = (s[i] == s[i + 1]);
    num[i][i + 1] = 2 + dp[i][i + 1];
  }
  for (__typeof(s.size() + 1) k = 3; k < (s.size() + 1); ++k) {
    for (__typeof(s.size() + 1 - k) i = 0; i < (s.size() + 1 - k); ++i) {
      int j = static_cast<int>(i + k - 1);
      dp[i][j] = dp[i + 1][j - 1] & (s[i] == s[j]);
      num[i][j] = num[i][j - 1] + num[i + 1][j] - num[i + 1][j - 1] + dp[i][j];
    }
  }
  int q;
  cin >> q;
  for (__typeof(q) i = 0; i < (q); ++i) {
    int x, y;
    cin >> x >> y;
    printf("%d\n", num[x - 1][y - 1]);
  }
  return 0;
}
