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;
long long dp[5005][5005];
bool arr[5005][5005];
string SS;
long long solve(int l, int r) {
  if (l > r) return 0;
  if (dp[l][r] != -1) return dp[l][r];
  long long ret;
  if (arr[l][r])
    ret = 1;
  else
    ret = 0;
  ret += solve(l + 1, r);
  ret += solve(l, r - 1);
  ret -= solve(l + 1, r - 1);
  return dp[l][r] = ret;
}
int main() {
  cin >> SS;
  for (int i = 0; i < SS.size(); i++) {
    for (int j = i, k = i; j < SS.size() && k >= 0 && SS[j] == SS[k]; k--, j++)
      arr[k][j] = 1;
    for (int j = i + 1, k = i; j < SS.size() && k >= 0 && SS[j] == SS[k];
         k--, j++)
      arr[k][j] = 1;
  }
  memset(dp, -1, sizeof(dp));
  int q;
  scanf("%d", &q);
  for (int cs = 1; cs <= q; cs++) {
    int l, r;
    scanf("%d %d", &l, &r);
    l--;
    r--;
    long long ans = solve(l, r);
    printf("%lld\n", ans);
  }
}
