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;
const double eps = 0.000000000000001;
char s[5010];
int l;
bool is[5010][5010];
int dp[5010][5010];
int main() {
  scanf("%s", s + 1);
  l = strlen(s + 1);
  int i, j;
  for (i = 1; i <= l; i++) {
    is[i][1] = 1;
    dp[i][1] = 1;
    if (i < l) {
      is[i][2] = (s[i] == s[i + 1] ? 1 : 0);
    }
  }
  int len;
  for (len = 3; len <= l; len++) {
    for (i = 1; i <= l - len + 1; i++) {
      is[i][len] = ((is[i + 1][len - 2]) && (s[i] == s[i + len - 1]));
    }
  }
  for (len = 2; len <= l; len++) {
    for (i = 1; i <= l - len + 1; i++) {
      dp[i][len] = dp[i + 1][len - 1] + dp[i][len - 1] - dp[i + 1][len - 2] +
                   (is[i][len]);
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    assert(l <= r);
    printf("%d\n", dp[l][r - l + 1]);
  }
  return 0;
}
