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>
const double EPS = 1e-9;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int P = 29;
using namespace std;
char second[5010];
bool p[5010][5010];
int n;
int d[5010][5010];
int main() {
  gets(second);
  n = (int)strlen(second);
  for (int len = 1; len <= n; len++)
    for (int l = 0; l <= n - len; l++) {
      int r = l + len - 1;
      if (l == r)
        p[l][r] = 1;
      else if (l == r - 1)
        p[l][r] = (second[l] == second[r]);
      else
        p[l][r] = p[l + 1][r - 1] && (second[l] == second[r]);
    }
  for (int len = 1; len <= n; len++)
    for (int l = 0; l <= n - len; l++) {
      int r = l + len - 1;
      if (l == r)
        d[l][r] = 1;
      else if (l == r - 1)
        d[l][r] = (second[l] == second[r]) + 2;
      else
        d[l][r] = d[l + 1][r] + d[l][r - 1] - d[l + 1][r - 1] + p[l][r];
    }
  int q;
  cin >> q;
  for (int i = 0; i < (int)q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    printf("%d\n", d[l][r]);
    continue;
  }
  return 0;
}
