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 N = 5e3 + 10;
char s[N];
bool d1[N][N];
long long d2[N][N];
int main() {
  while (scanf("%s", s) != EOF) {
    int len = strlen(s);
    for (int i = 0; i < len; i++) {
      for (int j = 0; j < i; j++) {
        d1[i][j] = 1;
        d2[i][j] = 0;
      }
      d1[i][i] = 1;
      d2[i][i] = 1;
    }
    for (int d = 2; d <= len; d++)
      for (int l = 0; l + d - 1 < len; l++) {
        int r = l + d - 1;
        d1[l][r] = (d1[l + 1][r - 1] && s[l] == s[r]);
      }
    for (int d = 2; d <= len; d++)
      for (int l = 0; l + d - 1 < len; l++) {
        int r = l + d - 1;
        d2[l][r] = d2[l][r - 1] + d2[l + 1][r] - d2[l + 1][r - 1] + d1[l][r];
      }
    int q, x, y;
    scanf("%d", &q);
    while (q--) {
      scanf("%d%d", &x, &y);
      printf("%lld\n", d2[x - 1][y - 1]);
    }
  }
  return 0;
}
