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(false), cin.tie(0);
  long long t = 1, m, i, j, k, u, to;
  while (t--) {
    string s;
    cin >> s;
    int n = s.length();
    vector<vector<int>> dp(n, vector<int>(n)), pal(n, vector<int>(n));
    for (i = 0; i < n; i++) {
      dp[i][i] = pal[i][i] = 1;
      if (i + 1 < n) {
        dp[i][i + 1] = 2;
        dp[i][i + 1] += pal[i][i + 1] = s[i] == s[i + 1];
      }
    }
    for (int len = 3; len <= n; len++) {
      for (i = 0; i <= n - len; i++) {
        j = i + len - 1;
        if (s[i] == s[j]) {
          pal[i][j] = pal[i + 1][j - 1];
        }
        dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1];
        dp[i][j] += pal[i][j];
      }
    }
    int q;
    cin >> q;
    while (q--) {
      cin >> u >> to;
      u--, to--;
      cout << dp[u][to] << '\n';
    }
  }
}
