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 v[5005][5005], dp[5005][5005];
char s[5005];
int main() {
  int L, T, l, i, j, left, right, m, n;
  scanf("%s", s);
  L = strlen(s);
  for (i = 0; i < L; i++) dp[i][i] = 1, v[i][i] = 1, v[i + 1][i] = 1;
  for (l = 1; l < L; l++) {
    for (i = 0; i < L - l; i++) {
      left = i;
      right = i + l;
      if (s[left] == s[right] && v[left + 1][right - 1]) v[left][right] = 1;
      dp[left][right] = dp[left + 1][right] + dp[left][right - 1] +
                        v[left][right] - dp[left + 1][right - 1];
    }
  }
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &m, &n);
    printf("%d\n", dp[m - 1][n - 1]);
  }
}
