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;
char s[5100];
int memo[5010][5010];
int v[5010][5010];
int n;
int main() {
  scanf("%s", s);
  n = strlen(s);
  memset(v, 0, sizeof(v));
  memset(memo, 0, sizeof(memo));
  for (int i = 1; i <= n; i++) {
    memo[i][i] = 1;
  }
  for (int i = n; i >= 1; i--) {
    for (int j = i + 1; j <= n; j++) {
      memo[i][j] = 0;
      if (s[i - 1] != s[j - 1]) continue;
      if (i + 1 > j - 1) {
        memo[i][j] = 1;
        continue;
      }
      if (memo[i + 1][j - 1]) {
        memo[i][j] = 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      v[i][j] = v[i - 1][j] + v[i][j - 1] - v[i - 1][j - 1];
      if (memo[i][j]) {
        v[i][j]++;
      }
    }
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < (q); i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    int res = v[b][b] - v[b][a - 1] - v[a - 1][b] + v[a - 1][a - 1];
    printf("%d\n", res);
  }
  return 0;
}
