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 OO = 1e8;
const double EPS = (1e-7);
int dcmp(double x, double y) { return fabs(x - y) <= EPS ? 0 : x < y ? -1 : 1; }
string s;
long long dp[5005][5005];
bool palin[5005][5005];
int n;
long long countP(int i, int j) {
  if (i > j) return 0;
  long long& ret = dp[i][j];
  if (ret != -1) {
    return ret;
  }
  if (i == j) {
    palin[i][j] = 1;
    return ret = 1;
  }
  ret = 0;
  ret += countP(i + 1, j);
  ret += countP(i, j - 1);
  ret -= countP(i + 1, j - 1);
  if (s[i] == s[j] && (palin[i + 1][j - 1] || i + 1 > j - 1)) {
    ret += 1;
    palin[i][j] = 1;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> s;
  n = s.length();
  memset(dp, -1, sizeof(dp));
  countP(0, n - 1);
  int q;
  cin >> q;
  int x, y;
  while (q--) {
    cin >> x >> y;
    cout << dp[x - 1][y - 1] << " ";
  }
}
