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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1e9 + 7;
long double EPS = 1e-12;
const int N = 5e3;
int dp[N][N];
int cnt[N][N];
string s;
int REC(int l, int r) {
  if (l == r) dp[l][r] = 1;
  if (l >= r) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  REC(l + 1, r), REC(l, r - 1);
  dp[l][r] = REC(l + 1, r - 1) & (s[l] == s[r]);
  return dp[l][r];
}
int REK(int l, int r) {
  if (l == r) {
    cnt[l][r] = 1;
    return 1;
  }
  if (l > r) return 0;
  if (cnt[l][r] != -1) return cnt[l][r];
  cnt[l][r] = REK(l + 1, r) + REK(l, r - 1) - REK(l + 1, r - 1) + dp[l][r];
  return cnt[l][r];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ifstream fin("ip.txt");
  ofstream fout("ip.txt");
  {
    cin >> s;
    memset(dp, -1, sizeof(dp)), memset(cnt, -1, sizeof(cnt));
    int n = s.length();
    REC(0, n - 1), REK(0, n - 1);
    int q;
    cin >> q;
    while (q--) {
      int L, R;
      cin >> L >> R;
      L--, R--;
      cout << cnt[L][R] << '\n';
    }
  }
}
