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;
long long int fast(long long int n, long long int a, long long int m) {
  if (n == 0)
    return 1;
  else if (n == 1)
    return a;
  else {
    long long int r = fast(n / 2, a, m);
    if (n % 2 == 0) {
      return (((r % m) * (r % m)) % m);
    } else {
      return (((((r % m) * a) % m) * r) % m);
    }
  }
}
int main() {
  string s;
  cin >> s;
  int n = s.length();
  bool dp[n + 1][n + 1];
  int p[n + 1][n + 1];
  int i, j;
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= n; j++) {
      dp[i][j] = false;
      if (i == j) {
        dp[i][j] = true;
      }
      p[i][j] = 0;
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = i - 1; j >= 1; j--) {
      if (i == j + 1) {
        if (s[i - 1] == s[j - 1]) dp[j][i] = true;
      } else {
        if (s[i - 1] == s[j - 1]) dp[j][i] |= dp[j + 1][i - 1];
      }
    }
  }
  for (j = 1; j <= n; j++) {
    for (i = j; i >= 1; i--) {
      if (i == j)
        p[i][j] = 1;
      else {
        p[i][j] = p[i + 1][j] + p[i][j - 1] - p[i + 1][j - 1] +
                  (dp[i][j] == true ? 1 : 0);
      }
    }
  }
  int Q;
  cin >> Q;
  int L, R;
  while (Q--) {
    scanf("%d%d", &L, &R);
    long long int ans = p[L][R];
    printf("%lld\n", ans);
  }
}
