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 ch[10000];
int p[5010][5010];
int dp[5010][5010];
int main(void) {
  scanf("%s", ch);
  int length = strlen(ch);
  for (int i = 0; i < length; ++i) {
    p[i][i] = 1;
    if (i >= 1 && ch[i - 1] == ch[i]) p[i - 1][i] = 1;
    if (i < length - 1 && ch[i + 1] == ch[i]) p[i][i + 1] = 1;
  }
  for (int i = 0; i < length; ++i)
    for (int l = 1; i - l >= 0 && i + 1 + l <= length; ++l) {
      if (p[i - l + 1][i + l - 1] && ch[i - l] == ch[i + l])
        p[i - l][i + l] = 1;
      else
        p[i - l][i + l] = 0;
      if (p[i - l + 1][i + l] && ch[i - l] == ch[i + l + 1])
        p[i - l][i + l + 1] = 1;
      else
        p[i - l][i + l + 1] = 0;
    }
  for (int i = 0; i < length; ++i) dp[i][i] = 1;
  for (int l = 2; l <= length; ++l)
    for (int i = 0; i + l - 1 < length; ++i)
      dp[i][i + l - 1] = dp[i][i + l - 2] + dp[i + 1][i + l - 1] -
                         dp[i + 1][i + l - 2] + p[i][i + l - 1];
  int t;
  scanf("%d", &t);
  while (t--) {
    int m, n;
    scanf("%d %d", &m, &n);
    printf("%d\n", dp[m - 1][n - 1]);
  }
  return 0;
}
