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 + 5;
long long dp[maxn][maxn], q;
bool is[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) {
    is[i][i] = 1;
    dp[i][i] = 1;
  }
  for (int sz = 2; sz < s.size() + 1; ++sz) {
    for (int i = 0; i < s.size() - sz + 1; ++i) {
      int j = i + sz - 1;
      is[i][j] = (s[i] == s[j] && (j - i == 1 || is[i + 1][j - 1]));
      if (j - i == 1)
        dp[i][j] = dp[i][i] + dp[j][j] + is[i][j];
      else
        dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1] + is[i][j];
    }
  }
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    ;
    x--;
    y--;
    cout << dp[x][y] << '\n';
  }
  return 0;
}
