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;
int Get() {
  char c;
  while (c = getchar(), c < '0' || c > '9')
    ;
  int X = 0;
  while (c >= '0' && c <= '9') {
    X = X * 10 + c - 48;
    c = getchar();
  }
  return X;
}
void Output(int X) {
  int Len = 0, Data[20];
  while (X) {
    Data[Len++] = X % 10;
    X /= 10;
  }
  if (!Len) Data[Len++] = 0;
  while (Len--) putchar(Data[Len] + 48);
  putchar('\n');
}
int main() {
  string S;
  cin >> S;
  int N = S.size();
  static int DP[5000][5000];
  for (int i = 0; i < N; i++) DP[i][i] = 1;
  for (int i = 1; i < N; i++) DP[i - 1][i] = (S[i - 1] == S[i]);
  for (int Len = 2; Len < N; Len++)
    for (int i = 0, j = Len; j < N; i++, j++)
      DP[i][j] = (S[i] == S[j]) && DP[i + 1][j - 1];
  for (int Len = 1; Len < N; Len++)
    for (int i = 0, j = Len; j < N; i++, j++)
      DP[i][j] += DP[i + 1][j] + DP[i][j - 1] - DP[i + 1][j - 1];
  int Query = Get();
  while (Query--) {
    int L = Get() - 1, R = Get() - 1;
    Output(DP[L][R]);
  }
  return 0;
}
