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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a * (b / (gcd(a, b))); }
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
inline long long toLL(string s) {
  long long v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
using namespace std;
char c[5001];
int N;
bool P[5001][5001];
bool S[5001][5001];
int DP[5001][5001];
bool solve(int i, int j) {
  if (i >= j) return 1;
  if (S[i][j]) return P[i][j];
  S[i][j] = 1;
  return P[i][j] = (c[i] == c[j]) && solve(i + 1, j - 1);
}
int main() {
  scanf("%s", c);
  N = strlen(c);
  memset(P, -1, sizeof(P));
  memset(S, 0, sizeof(S));
  memset(DP, 0, sizeof(0));
  for (int i = 0; i < N; i++) {
    for (int j = i; j < N; j++) {
      DP[i][j] = solve(i, j);
    }
  }
  int q;
  scanf("%d", &q);
  int l, r;
  for (int i = 0; i < N; i++) {
    for (int j = 1; j < N; j++) {
      DP[i][j] += DP[i][j - 1];
    }
  }
  for (int i = N - 1; i >= 0; i--) {
    for (int j = 0; j < N; j++) {
      DP[i][j] += DP[i + 1][j];
    }
  }
  while (q--) {
    scanf("%d%d", &l, &r);
    --l;
    --r;
    printf("%d\n", DP[l][r]);
  }
}
