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;
bool isPal[5009][5009];
long long int dp[5009][5009];
int main() {
  long long int i, j, t1, t2, t3, t4, l, m, ans, len, q;
  string s1;
  cin >> s1;
  l = s1.length();
  for (i = 0; i <= 5000; i++) {
    for (j = 0; j <= 5000; j++) {
      isPal[i][j] = false;
      dp[i][j] = 0;
    }
  }
  for (i = 0; i < l; i++) {
    dp[i][i] = 1;
    isPal[i][i] = true;
  }
  for (len = 2; len <= l; len++) {
    for (i = 0; i < l; i++) {
      if (i + len - 1 >= l) {
        break;
      }
      t1 = i;
      t2 = i + len - 1;
      if (len == 2) {
        isPal[t1][t2] = (s1[t1] == s1[t2]) ? true : false;
        dp[t1][t2] = dp[t1 + 1][t2] + dp[t1][t2 - 1] - dp[t1 + 1][t2 - 1] +
                     (isPal[t1][t2]);
      } else {
        if (s1[t1] == s1[t2] && isPal[t1 + 1][t2 - 1]) {
          isPal[t1][t2] = true;
        }
        dp[t1][t2] = dp[t1 + 1][t2] + dp[t1][t2 - 1] - dp[t1 + 1][t2 - 1] +
                     (isPal[t1][t2]);
      }
    }
  }
  scanf("%lld", &q);
  while (q--) {
    scanf("%lld %lld", &t1, &t2);
    t1--;
    t2--;
    ans = dp[t1][t2];
    printf("%lld\n", ans);
  }
  return 0;
}
