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;
int ceil(int a, int b) {
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
bool checkbit(int pos, int mask) { return mask & (1 << pos); }
int turnon(int pos, int mask) { return mask | (1 << pos); }
const int N = 5005;
string s;
int dp[N][N], pref[N][N];
int fun(int i, int j) {
  if (i > j) return 1;
  if (dp[i][j] != -1) return dp[i][j];
  int ans = 0;
  if (s[i] == s[j]) ans |= fun(i + 1, j - 1);
  return dp[i][j] = ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  memset(dp, -1, sizeof(dp));
  cin >> s;
  s = " " + s;
  int n = s.size();
  for (int i = 1; i < n; i++) {
    for (int j = i; j < n; j++) {
      int tmp = fun(i, j);
    }
  }
  for (int i = 1; i < n; i++) {
    pref[i][0] = 0;
    for (int j = i; j < n; j++) {
      pref[i][j] = pref[i][j - 1] + dp[i][j];
    }
  }
  for (int i = 1; i < n; i++) {
    for (int j = i - 1; j >= 1; j--) {
      pref[j][i] += pref[j + 1][i];
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    cout << pref[x][y] << "\n";
  }
}
