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;
const int maxn = 5001;
char s[maxn];
int ispal[maxn][maxn];
int prec[maxn][maxn];
int main() {
  while (scanf("%s", s) >= 1) {
    int n = strlen(s);
    memset(ispal, 0, sizeof(ispal));
    for (int len = 0; len <= n; len++) {
      for (int i = 0; i <= n - len; i++) {
        ispal[i][i + len] =
            (len <= 1 ? 1
                      : ispal[i + 1][i + len - 1] && (s[i] == s[i + len - 1]));
        prec[i][i + len] = (len ? prec[i][i + len - 1] : 0) +
                           (len ? prec[i + 1][i + len] : 0) -
                           (len >= 2 ? prec[i + 1][i + len - 1] : 0) +
                           (len ? ispal[i][i + len] : 0);
      }
    }
    int m;
    scanf("%d", &m);
    for (int i = 0; i < m; i++) {
      int l, r;
      scanf("%d%d", &l, &r);
      --l;
      printf("%d\n", prec[l][r]);
    }
  }
  return 0;
}
