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;
const long long dx[4] = {-1, 1, 0, 0};
const long long dy[4] = {0, 0, -1, 1};
const long long MAX = 5e3 + 5;
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long n, m, ans, k, Q;
string second;
bool pal[MAX][MAX];
long long cnt[MAX][MAX];
void solve() {
  for (int i = 0; i < n; ++i) pal[i][i] = cnt[i][i] = true;
  for (int i = 0; i < n - 1; ++i) {
    if (second[i] == second[i + 1]) pal[i][i + 1] = true, cnt[i][i + 1] = 1;
    cnt[i][i + 1] += 2;
  }
  for (int len = 3; len <= n; ++len)
    for (int i = 0; i < n - len + 1; ++i) {
      int j = i + len - 1;
      if (second[i] == second[j] && pal[i + 1][j - 1])
        pal[i][j] = true;
      else
        pal[i][j] = false;
      cnt[i][j] = pal[i][j] + cnt[i][j - 1] + cnt[i + 1][j] - cnt[i + 1][j - 1];
    }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> second;
  n = second.length();
  solve();
  cin >> Q;
  while (Q--) {
    int lo, hi;
    cin >> lo >> hi;
    --lo;
    --hi;
    cout << cnt[lo][hi] << '\n';
  }
  return 0;
}
