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;
bool c[5005][5005] = {0};
int dp[5005][5005] = {0};
int main() {
  char s[5005];
  scanf("%s", s);
  int t = strlen(s);
  int i, j, k;
  for (i = 0; i < t; i++) {
    for (j = 0;; j++) {
      if ((i - j) < 0 || (i + j) >= t) break;
      if (s[i - j] == s[i + j])
        c[i - j + 1][i + j + 1] = true;
      else
        break;
    }
    for (j = 0;; j++) {
      if ((i - j - 1) < 0 || (i + j) >= t) break;
      if (s[i - j - 1] == s[i + j])
        c[i - j][i + j + 1] = true;
      else
        break;
    }
  }
  int q;
  scanf("%d", &q);
  for (i = 1; i <= t; i++) {
    dp[i][i] = 1;
  }
  for (i = 2; i <= t; i++)
    for (j = 1; j + i - 1 <= t; j++) {
      k = j + i - 1;
      dp[j][k] = dp[j + 1][k] + dp[j][k - 1] - dp[j + 1][k - 1] + c[j][k];
    }
  while (q--) {
    int aa, bb;
    scanf("%d%d", &aa, &bb);
    printf("%d\n", dp[aa][bb]);
  }
}
