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 s[5010];
bool flag[5010][5010];
int dp[5010][5010];
void init() {
  memset(flag, 0, sizeof(flag));
  memset(dp, 0, sizeof(dp));
  s[0] = '#';
  int n = strlen(s);
  for (int i = 1; i <= n; i++) flag[i][i] = 1;
  for (int len = 2; len <= n; len++) {
    for (int i = 1; i + len - 1 <= n; i++) {
      int j = i + len - 1;
      if (s[i] != s[j]) continue;
      if (len == 2 || flag[i + 1][j - 1]) flag[i][j] = 1;
    }
  }
  for (int len = 1; len <= n; len++)
    for (int i = 1; i + len - 1 <= n; i++)
      dp[i][i + len - 1] = dp[i + 1][i + len - 1] + dp[i][i + len - 2] -
                           dp[i + 1][i + len - 2] + flag[i][i + len - 1];
}
int main() {
  int n;
  while (~scanf("%s%d", s + 1, &n)) {
    init();
    int x, y;
    while (n--) {
      scanf("%d%d", &x, &y);
      printf("%d\n", dp[x][y]);
    }
  }
  return 0;
}
