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;
int power(int b, int e, int m) {
  int p = 1;
  while (e > 0) {
    if (e & 1) p = (p * b) % m;
    e = e >> 1;
    b = (b * b) % m;
  }
  return p;
}
char s[5005];
int dp[5005][5005], isPal[5005][5005];
int calcIsPal(int l, int r) {
  if (l > r + 1) {
    return 0;
  }
  if (l == r + 1) {
    return 1;
  }
  if (isPal[l][r] != -1) {
    return isPal[l][r];
  }
  if (l == r) {
    isPal[l][l] = 1;
  } else {
    isPal[l][r] = (s[l] == s[r]) && calcIsPal(l + 1, r - 1);
  }
  return isPal[l][r];
}
int solve(int l, int r) {
  if (l > r) {
    return 0;
  }
  if (dp[l][r] != -1) {
    return dp[l][r];
  }
  if (l == r) {
    dp[l][l] = 1;
  } else {
    dp[l][r] = calcIsPal(l, r) - solve(l + 1, r - 1) + solve(l + 1, r) +
               solve(l, r - 1);
  }
  return dp[l][r];
}
int main() {
  int n, i, t, j, h, k = 0, ans = 0, h1, h2, q, l, r;
  scanf("%s", s);
  scanf("%d", &q);
  n = strlen(s);
  memset(dp, -1, sizeof(dp));
  memset(isPal, -1, sizeof(isPal));
  solve(0, n - 1);
  while (q--) {
    scanf("%d", &l);
    scanf("%d", &r);
    printf("%d\n", dp[l - 1][r - 1]);
  }
  return 0;
}
