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 ispal[5010][5010];
int dp[5010][5010];
char s[5010];
int main() {
  scanf("%s", s);
  memset(dp, 0, sizeof(dp));
  int len = strlen(s), i, start, l;
  for (i = 1; i <= len; ++i) {
    dp[i][i] = 1;
    ispal[i][i] = 1;
  }
  if (len >= 2) {
    for (start = 1; start + 1 <= len; ++start)
      if (s[start - 1] == s[start]) ispal[start][start + 1] = 1;
  }
  for (l = 3; l <= len; ++l)
    for (start = 1; start + l - 1 <= len; ++start) {
      ispal[start][start + l - 1] = (ispal[start + 1][start + l - 2] &
                                     (bool)(s[start - 1] == s[start + l - 2]));
    }
  for (l = 2; l <= len; ++l)
    for (start = 1; start + l - 1 <= len; ++start) {
      dp[start][start + l - 1] =
          dp[start + 1][start + l - 1] + dp[start][start + l - 2] -
          dp[start + 1][start + l - 2] + (int)(ispal[start][start + l - 1]);
    }
  int q, r;
  scanf("%d", &q);
  for (i = 0; i < q; ++i) {
    scanf("%d%d", &l, &r);
    printf("%d\n", dp[l][r]);
  }
  return 0;
}
