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;
template <class T1, class T2>
int upmin(T1 &x, T2 v) {
  if (x > v) {
    x = v;
    return 1;
  }
  return 0;
}
template <class T1, class T2>
int upmax(T1 &x, T2 v) {
  if (x < v) {
    x = v;
    return 1;
  }
  return 0;
}
const int INF = 0x3f3f3f3f;
string seq;
void init() { cin >> seq; }
bool is_pal[5000 + 1][5000 + 1];
int dp[5000 + 1][5000 + 1];
void preprocess() {
  for (int i = 0; i < seq.size(); ++i) {
    is_pal[i][i] = dp[i][i] = 1;
  }
  for (int i = 0; i + 1 < seq.size(); ++i) {
    if (seq[i] == seq[i + 1]) {
      is_pal[i][i + 1] = dp[i][i + 1] = 1;
    }
  }
  for (int len = 3; len <= seq.size(); ++len) {
    for (int i = 0; i + len <= seq.size(); ++i) {
      if (seq[i] == seq[i + len - 1]) {
        is_pal[i][i + len - 1] = is_pal[i + 1][i + len - 2];
      }
    }
  }
  for (int len = 2; len <= seq.size(); ++len) {
    for (int i = 0; i + len <= seq.size(); ++i) {
      dp[i][i + len - 1] = dp[i][i + len - 2] + dp[i + 1][i + len - 1] -
                           dp[i + 1][i + len - 2] + is_pal[i][i + len - 1];
    }
  }
}
void solve() {
  int Q;
  cin >> Q;
  for (int kase = 0; kase < Q; ++kase) {
    int ql, qr;
    scanf("%d%d", &ql, &qr);
    printf("%d\n", dp[ql - 1][qr - 1]);
  }
}
signed main() {
  init();
  preprocess();
  solve();
  return 0;
}
