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;
long long t, n, u, v, m, q, r, ql, qr, k, l, s, x, y, w, h, c, z, d, b;
const int N = 5e3 + 1;
const long long mod = 1e9 + 7;
const long long INF = 1LL << 62LL;
const bool JUDGE = false;
bool pal[N][N];
int dp[N][N];
string str;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (JUDGE) {
    freopen("cricket.in", "r", stdin);
    freopen("cricket.out", "w", stdout);
  }
  cin >> str >> q;
  n = str.size();
  for (int i = 0; i < str.size(); ++i) {
    pal[i][i] = true;
    if (i < (int)str.size() - 1) pal[i][i + 1] = (str[i] == str[i + 1]);
    dp[i][i] = 1;
  }
  for (int len = 3; len <= str.size(); ++len) {
    for (int i = 0; i < str.size(); ++i) {
      if (i + len - 1 >= str.size()) continue;
      pal[i][i + len - 1] =
          pal[i + 1][i + len - 2] & (str[i] == str[i + len - 1]);
    }
  }
  for (int i = 2; i <= str.size(); ++i) {
    for (int j = 0; j < str.size(); ++j) {
      if (j + i - 1 >= str.size()) continue;
      dp[j][j + i - 1] = dp[j + 1][j + i - 1] + dp[j][j + i - 2] -
                         dp[j + 1][j + i - 2] + pal[j][j + i - 1];
    }
  }
  while (q--) {
    cin >> l >> r;
    cout << dp[l - 1][r - 1] << '\n';
  }
  return 0;
}
