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;
using ll = long long;
double eps = 0.0000001;
int p1_ = 63;
int m1_ = 1000000009;
void get_h(string &s, vector<int> &h_pref, vector<int> &pow_p,
           long long p = p1_, int m = m1_) {
  pow_p[0] = 1;
  int n = pow_p.size();
  for (int i = 1; i < n; i++) {
    pow_p[i] = pow_p[i - 1] * p % m;
  }
  h_pref[0] = s[0] - 'a' + 1;
  for (int i = 1; i < n; i++) {
    h_pref[i] = (h_pref[i - 1] + (long long)pow_p[i] * (s[i] - 'a' + 1)) % m;
  }
}
vector<int> h1;
vector<int> h2;
vector<int> p1;
vector<int> p2;
int n;
bool is_p(int l, int r) {
  long long hsh;
  if (l != 0) {
    hsh = h1[r] - h1[l - 1];
  } else {
    hsh = h1[r];
  }
  if (hsh < 0) {
    hsh += m1_;
  }
  hsh = (hsh * p1[n - 1 - r]) % m1_;
  int d = r - l;
  l = n - 1 - r;
  r = l + d;
  long long hsh2;
  if (l != 0) {
    hsh2 = h2[r] - h2[l - 1];
  } else {
    hsh2 = h2[r];
  }
  if (hsh2 < 0) {
    hsh2 += m1_;
  }
  hsh2 = (hsh2 * p2[n - 1 - r]) % m1_;
  return hsh == hsh2;
}
int main() {
  string s;
  cin >> s;
  n = s.size();
  h1.resize(n);
  h2.resize(n);
  p1.resize(n);
  p2.resize(n);
  get_h(s, h1, p1);
  reverse(s.begin(), s.end());
  get_h(s, h2, p2);
  vector<vector<int>> dp(5050, vector<int>(5050));
  for (int i = 0; i < n; i++) {
    dp[i][i] = 1;
  }
  for (int l = 2; l <= n; l++) {
    for (int i = 0; i <= n - l; i++) {
      int j = i + l - 1;
      dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1] + is_p(i, j);
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    printf("%d\n", dp[a][b]);
  }
  return 0;
}
