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 int N = 20010;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
const double e = 2.718281828459046;
using namespace std;
int dp[5010][5010];
char s[5010];
unsigned long long h1[5010], h2[5010];
vector<int> vt[5010];
unsigned long long p[5010];
int len;
unsigned long long f1(int l, int r) {
  if (l == 0) return h1[r] * (p[len - r - 1]);
  return (h1[r] - h1[l - 1]) * (p[len - r - 1]);
}
unsigned long long f2(int l, int r) { return (h2[l] - h2[r + 1]) * p[l]; }
int t[5010];
int main() {
  scanf("%s", s);
  len = strlen(s);
  p[0] = 1;
  for (int i = 1; i <= len; i++) p[i] = p[i - 1] * 31;
  for (int i = 0; s[i]; i++) {
    if (i == 0)
      h1[i] = s[0] - 'a';
    else
      h1[i] = h1[i - 1] + p[i] * (s[i] - 'a');
  }
  for (int i = len - 1, j = 0; i >= 0; j++, i--)
    h2[i] = h2[i + 1] + p[j] * (s[i] - 'a');
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++)
      if (f1(i, j) == f2(i, j)) vt[i].push_back(j);
  for (int k = 0; k < len; k++)
    for (int i = 0; i + k < len; i++) {
      if (k == 0)
        dp[i][i + k] = 1;
      else {
        while (t[i] < (int((vt[i]).size()))) {
          if (vt[i][t[i]] <= i + k)
            t[i]++;
          else
            break;
        }
        dp[i][i + k] = dp[i + 1][i + k] + t[i];
      }
    }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--, r--;
    printf("%d\n", dp[l][r]);
  }
  return 0;
}
