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;
struct Query {
  int lo, hi, id;
} q[1000111];
int n, tree[5005], res[1000111];
string s;
void update(int i) {
  while (i > 0) {
    tree[i]++;
    i -= (-i & i);
  }
}
int get(int i) {
  int res = 0;
  while (i <= s.length()) {
    res += tree[i];
    i += (-i & i);
  }
  return res;
}
bool cmp(const Query &a, const Query &b) { return (a.hi < b.hi); }
int main() {
  cin >> s;
  scanf("%d", &n);
  for (int i = 0, _n = (n); i < _n; i++) {
    scanf("%d %d", &q[i].lo, &q[i].hi);
    q[i].id = i;
  }
  sort(q, q + n, cmp);
  int k = 1;
  for (int i = 0, _n = (n); i < _n; i++) {
    while (k <= q[i].hi && k <= s.length()) {
      unsigned long long ha = 0, hb = 0, mul = 1;
      for (int j = (k), _b = (1); j >= _b; j--) {
        ha = ha * 9875321 + s[j - 1];
        hb = hb + mul * s[j - 1];
        mul *= 9875321;
        if (ha == hb) update(j);
      }
      k++;
    }
    res[q[i].id] = get(q[i].lo);
  }
  for (int i = 0, _n = (n); i < _n; i++) printf("%d\n", res[i]);
  return 0;
}
