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>
const int MAXN = 5050;
char word[MAXN];
bool palin[MAXN][MAXN];
int soma[MAXN][MAXN];
void dp(int n) {
  for (int i = 1; i <= n; ++i) {
    palin[i][i] = true;
  }
  for (int l = 2; l <= n; ++l) {
    for (int i = 1; i <= n - l + 1; ++i) {
      int j = i + l - 1;
      if (l == 2)
        palin[i][j] = (word[i] == word[j]);
      else
        palin[i][j] = (word[i] == word[j] && palin[i + 1][j - 1]);
    }
  }
}
void acum(int n) {
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      soma[i][j] = (int)palin[i][j] + soma[i - 1][j] + soma[i][j - 1] -
                   soma[i - 1][j - 1];
    }
  }
}
int query(int a, int b) {
  return soma[b][b] - soma[b][a - 1] - soma[a - 1][b] + soma[a - 1][a - 1];
}
int main() {
  scanf(" %s", &word[1]);
  int n = strlen(&word[1]);
  dp(n);
  acum(n);
  int q;
  scanf(" %d", &q);
  int a, b;
  for (int i = 0; i < q; ++i) {
    scanf(" %d %d", &a, &b);
    printf("%d\n", query(a, b));
  }
  return 0;
}
