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 maxn = 5e3 + 10;
bool f[maxn][maxn];
int dp[maxn][maxn];
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    int j = 0;
    while (i - j >= 0 and i + j < s.size()) {
      if (s[i - j] != s[i + j]) break;
      f[i - j][i + j] = true;
      j++;
    }
  }
  for (int i = 1; i < s.size(); i++) {
    int j = 0;
    while (i - 1 - j >= 0 and i + j < s.size()) {
      if (s[i - 1 - j] != s[i + j]) break;
      f[i - 1 - j][i + j] = true;
      j++;
    }
  }
  for (int i = 0; i < s.size(); i++) dp[i][i] = 1;
  for (int i = 0; i + 1 < s.size(); i++)
    dp[i][i + 1] = (s[i] == s[i + 1] ? 3 : 2);
  for (int i = 2; i < s.size(); i++)
    for (int j = 0; j + i < s.size(); j++)
      dp[j][j + i] = dp[j][j + i - 1] + dp[j + 1][j + i] -
                     dp[j + 1][j + i - 1] + f[j][j + i];
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    cout << dp[l - 1][r - 1] << '\n';
  }
  return 0;
}
