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 mod = 1e9 + 7;
long long dx[] = {-1, 0, 1, 0};
long long dy[] = {0, -1, 0, 1};
long long dp[5005][5005];
bool p[5005][5005];
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  long long n = s.size();
  memset(dp, 0, sizeof(dp));
  memset(p, 0, sizeof(p));
  for (long long i = 0; i < n; i++) {
    dp[i + 1][i + 1] = p[i + 1][i + 1] = 1;
  }
  for (long long len = 2; len <= n; len++) {
    for (long long i = 1; i <= n - len + 1; i++) {
      long long j = i + len - 1;
      if (i + 1 > j - 1) {
        if (s[i - 1] == s[j - 1]) {
          p[i][j] = 1;
        }
      } else if (i + 1 <= j - 1 && s[i - 1] == s[j - 1] && p[i + 1][j - 1])
        p[i][j] = 1;
      else
        p[i][j] = 0;
      dp[i][j] = dp[i + 1][j] + dp[i][j - 1] + p[i][j] - dp[i + 1][j - 1];
    }
  }
  long long q;
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    cout << dp[l][r] << "\n";
  }
}
