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 q;
char s[5050];
bool f[5000][5000];
int g[5000][5000];
int main() {
  scanf("%s", s);
  int n = strlen(s);
  for (int i = 0; i < n; i++) {
    for (int j = i; j >= 0; j--) {
      if (i - j == 0)
        f[j][i] = 1;
      else if (i - j == 1 && s[i] == s[j])
        f[j][i] = 1;
      else if (f[j + 1][i - 1] && s[i] == s[j])
        f[j][i] = 1;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j >= 0; j--) {
      if (i - j == 0)
        g[j][i] = 1;
      else {
        g[j][i] = g[j + 1][i] + g[j][i - 1];
        if (i - j != 1) g[j][i] -= g[j + 1][i - 1];
        if (f[j][i]) g[j][i]++;
      }
    }
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", g[l - 1][r - 1]);
  }
  return 0;
}
