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 p[5005];
long long l[5005], r[5005];
int x[5005][5005];
long long inv[5005];
long long int powmod(long long int base, long long int exponent) {
  if (exponent == 0) return 1;
  long long int temp = powmod(base, exponent / 2);
  temp = temp % 1000000007;
  if (exponent % 2 == 0)
    return (temp * temp) % 1000000007;
  else
    return ((base % 1000000007) * ((temp * temp) % 1000000007)) % 1000000007;
}
string s;
void h(int n) {
  p[0] = 1;
  for (int i = 1; i <= 5005; i++) p[i] = (p[i - 1] * 29) % 1000000007;
  for (int i = 1; i <= n; i++) {
    l[i] = (l[i - 1] + (p[i] * (s[i] - 'a')) % 1000000007) % 1000000007;
  }
  for (int i = n; i >= 1; i--) {
    r[i] = (r[i + 1] + (p[n - i + 1] * (s[i] - 'a')) % 1000000007) % 1000000007;
  }
}
int main() {
  cin >> s;
  s = '&' + s;
  int n = s.size();
  h(n);
  for (int i = 0; i <= n; i++) {
    inv[i] = powmod(p[i], 1000000007 - 2);
  }
  for (int j = 1; j <= n; j++) {
    for (int i = j; i >= 1; i--) {
      int temp = 0;
      long long c =
          (1ll * ((l[j] - l[i - 1] + 1000000007) % 1000000007) * inv[i - 1]) %
          1000000007;
      long long d =
          (1ll * ((r[i] - r[j + 1] + 1000000007) % 1000000007) * inv[n - j]) %
          1000000007;
      if (c == d) {
        temp = 1;
      }
      x[j][i] = x[j][i + 1] + temp;
    }
  }
  for (int j = 1; j <= n; j++) {
    for (int i = j; i >= 1; i--) {
      x[j][i] = x[j - 1][i] + x[j][i];
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    scanf("%d", &l);
    scanf("%d", &r);
    printf("%d\n", x[r][l]);
  }
  return 0;
}
