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 n, q, f[5005][5005], l, r;
bool vis[5005][5005];
char s[1000005];
signed main() {
  cin >> s + 1;
  n = strlen(s + 1);
  for (long long i = 1; i <= n; i++) {
    vis[i][i] = 1;
    f[i][i] = 1;
    for (long long p1 = i - 1, p2 = i + 1; p1 >= 1 && p2 <= n; p1--, p2++) {
      if (s[p1] == s[p2])
        vis[p1][p2] = 1;
      else
        break;
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long p1 = i - 1, p2 = i; p1 >= 1 && p2 <= n; p1--, p2++) {
      if (s[p1] == s[p2])
        vis[p1][p2] = 1;
      else
        break;
    }
  }
  for (long long i = 2; i <= n; i++) {
    for (long long l = 1, r = i; r <= n; l++, r++)
      f[l][r] = f[l + 1][r] + f[l][r - 1] - f[l + 1][r - 1] + vis[l][r];
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &l, &r);
    printf("%lld\n", f[l][r]);
  }
}
