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;
constexpr long long MOD = 1000000009;
constexpr int P1 = 317;
constexpr int N = 5010;
long long hash1, hash_all, t, t2, l0, l1, l2, llp1 = 1;
int dp[N][N], h[N], q, u, v;
string n;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  l0 = (int)n.size();
  for (int i = 1; i <= l0; i++) {
    t += n[i - 1];
    t *= P1;
    t %= MOD;
    hash1 = t;
    llp1 *= P1;
    llp1 %= MOD;
    for (int j = 0; j <= l0 - i; j++) {
      h[j] = hash1;
      hash1 -= llp1 * n[j];
      hash1 = ((hash1 % MOD) + MOD) % MOD;
      hash1 += n[j + i];
      hash1 *= P1;
      hash1 %= MOD;
    }
    t2 += n[n.size() - i];
    t2 *= P1;
    t2 %= MOD;
    hash1 = t2;
    for (int j = l0 - i; j >= 0; j--) {
      if (h[j] == hash1) dp[j][j + i - 1] = true;
      hash1 -= llp1 * n[j + i - 1];
      hash1 = ((hash1 % MOD) + MOD) % MOD;
      if (j - 1 >= 0) hash1 += n[j - 1];
      hash1 *= P1;
      hash1 %= MOD;
    }
  }
  for (int l = 1; l <= l0; l++) {
    for (int i = 0; i < l0; i++) {
      dp[i][i + l - 1] +=
          dp[i + 1][i + l - 1] + dp[i][i + l - 2] - dp[i + 1][i + l - 2];
      dp[i][i + l - 1] %= MOD;
    }
  }
  for (int i = 0; i < q; i++) {
    cin >> u >> v;
    cout << dp[u - 1][v - 1] << '\n';
  }
  return 0;
}
