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;
bool isP[5002][5002];
int dp[5002][5002];
char str[5002];
void isPalin() {
  int len = strlen(str);
  for (int l = 0; l < len; l++) {
    for (int i = 0; i < len - l; i++) {
      int j = i + l;
      if (i == j)
        isP[i][i] = 1;
      else if (l == 1 && str[i] == str[j])
        isP[i][j] = 1;
      else if (l == 1 && str[i] != str[j])
        isP[i][j] = 0;
      else {
        if (str[i] == str[j] && isP[i + 1][j - 1] == 1)
          isP[i][j] = 1;
        else
          isP[i][j] = 0;
      }
    }
  }
}
void func() {
  int len = strlen(str);
  for (int l = 0; l < len; l++) {
    for (int i = 0; i < len - l; i++) {
      int j = i + l;
      if (i == j)
        dp[i][i] = 1;
      else if (l == 1 && str[i] == str[j])
        dp[i][j] = 3;
      else if (l == 1 && str[i] != str[j])
        dp[i][j] = 2;
      else {
        dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1] +
                   (str[i] == str[j] && isP[i + 1][j - 1]);
      }
    }
  }
}
int main() {
  int n, i, j;
  scanf("%s", &str);
  scanf("%d", &n);
  isPalin();
  func();
  while (n--) {
    scanf("%d %d", &i, &j);
    cout << dp[i - 1][j - 1] << "\n";
  }
  return 0;
}
