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;
const long long M = 1e9 + 9;
const long long base = 347;
const int N = 5007;
long long pb[N];
void pre() {
  pb[0] = 1;
  for (int i = 1; i < N; i++) pb[i] = (pb[i - 1] * base) % M;
}
vector<long long> hashList(string s) {
  int n = s.size();
  vector<long long> ans(n + 1);
  ans[0] = 0;
  for (int i = 1; i <= n; i++) ans[i] = (ans[i - 1] * base + s[i - 1]) % M;
  return ans;
}
long long Hash(vector<long long> &hashlist, int l, int r) {
  int len = (r - l + 1);
  return ((hashlist[r] - hashlist[l - 1] * pb[len]) % M + M) % M;
}
int dp[N][N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  pre();
  string s, t;
  cin >> s;
  t = s;
  int n = s.size();
  reverse(t.begin(), t.end());
  vector<long long> h = hashList(s);
  vector<long long> rh = hashList(t);
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      long long front = Hash(h, i, j);
      long long back = Hash(rh, n + 1 - j, n + 1 - i);
      if (front == back) dp[i][j] = 1;
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      dp[i][j] += dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1];
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << dp[r][r] - dp[l - 1][r] - dp[r][l - 1] + dp[l - 1][l - 1] << "\n";
  }
}
