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;
#pragma comment(linker, "/STACK:64000000")
long long p = 59;
long long deg[5005];
long long dh[5005], udh[5005];
int polindrom[5005][5005];
int dp[5005][5005];
void createDeg() {
  deg[0] = 1;
  for (int i = 1; i < 5005; i++) deg[i] = deg[i - 1] * p;
}
void createDirectHash(string s) {
  for (int i = 0; i < s.size(); i++) {
    dh[i] = s[i] * deg[i];
    if (i) dh[i] += dh[i - 1];
  }
}
void createUndirectHash(string s) {
  for (int i = s.size() - 1; i >= 0; i--) {
    udh[i] = udh[i + 1] + s[i] * deg[s.size() - 1 - i];
  }
}
int main() {
  string s;
  cin >> s;
  createDeg();
  createDirectHash(s);
  createUndirectHash(s);
  for (int l = 0; l < s.size(); l++) {
    for (int r = l; r < s.size(); r++) {
      long long hd = dh[r];
      if (l) hd -= dh[l - 1];
      long long hud = udh[l] - udh[r + 1];
      if (l > (s.size() - 1 - r)) {
        hud *= deg[l - (s.size() - 1 - r)];
      } else {
        hd *= deg[s.size() - 1 - r - l];
      }
      if (hud == hd) {
        polindrom[l + 1][r + 1]++;
      }
    }
  }
  for (int i = 1; i < 5005; i++)
    for (int j = 1; j < 5005; j++)
      dp[i][j] =
          dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + polindrom[i][j];
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    cout << dp[r][r] - dp[l - 1][r] - dp[r][l - 1] + dp[l - 1][l - 1] << endl;
  }
  return 0;
}
