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;
long long powmod(long long a, long long b) {
  long long res = 1;
  if (a >= 1000000007) a %= 1000000007;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    if (res >= 1000000007) res %= 1000000007;
    a = a * a;
    if (a >= 1000000007) a %= 1000000007;
  }
  return res;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int n;
string x;
int dp[5001][5001];
bool isPalindrome[5001][5001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> x;
  cin >> n;
  int z = x.size() - 1;
  int i, j;
  for (i = 0; i <= 5000; i++) {
    for (j = 0; j <= 5000; j++) {
      dp[i][j] = 0;
      isPalindrome[i][j] = false;
    }
  }
  for (i = 0; i <= z; i++) {
    isPalindrome[i][i] = true;
    dp[i][i] = 1;
    if (i != z) {
      isPalindrome[i + 1][i] = true;
    }
  }
  for (int l = 1; l <= z; l++) {
    i = 0;
    j = l;
    while (i <= z && j <= z) {
      isPalindrome[i][j] = (isPalindrome[i + 1][j - 1]) & (x[i] == x[j]);
      dp[i][j] =
          dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1] + (isPalindrome[i][j]);
      i++;
      j++;
    }
  }
  for (i = 1; i <= n; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << dp[l][r] << "\n";
  }
  return 0;
}
