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;
char str[5001];
int q;
int pal[5001][5001], pre[5001][5001];
int main() {
  scanf("%5000s%d", str, &q);
  for (int i = 1; i <= 5000; i++) {
    pal[i][i] = 1;
  }
  for (int i = 1; i < 5000; i++) {
    if (str[i - 1] == str[i]) {
      pal[i][i + 1] = 1;
    }
  }
  for (int i = 3; i <= 5000; i++) {
    for (int j = 1; j <= 5000; j++) {
      if (j + i - 1 > 5000) {
        break;
      }
      pal[j][j + i - 1] =
          (pal[j + 1][j + i - 2] && str[j - 1] == str[j + i - 2]);
    }
  }
  for (int i = 1; i <= 5000; i++) {
    for (int j = 1; j <= 5000; j++) {
      pre[i][j] = pal[i][j] + pre[i - 1][j] + pre[i][j - 1] - pre[i - 1][j - 1];
    }
  }
  for (int i = 0; i < q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n",
           pre[r][r] - pre[r][l - 1] - pre[l - 1][r] + pre[l - 1][l - 1]);
  }
  return 0;
}
