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;
const int N = 5001;
char str[N];
string s;
const int B = 29;
const int MOD = 1e9 + 9;
long long power[N];
long long preHash[N], sufHash[N];
int dp[N][N];
bool isPalindrome(int l, int r) {
  int n = s.length();
  long long subHash1 = preHash[r];
  if (l > 0) {
    subHash1 =
        (subHash1 - (preHash[l - 1] * power[r - l + 1]) % MOD + MOD) % MOD;
  }
  long long subHash2 = sufHash[l];
  if (r < n - 1) {
    subHash2 =
        (subHash2 - (sufHash[r + 1] * power[r - l + 1]) % MOD + MOD) % MOD;
  }
  return subHash1 == subHash2;
}
int solve(int l, int r) {
  if (l == r) {
    return 1;
  }
  if (l > r) {
    return 0;
  }
  int& ans = dp[l][r];
  if (ans != -1) {
    return ans;
  }
  ans = isPalindrome(l, r);
  ans += solve(l + 1, r);
  ans += solve(l, r - 1);
  ans -= solve(l + 1, r - 1);
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  scanf("%s", str);
  s = str;
  int n = s.length();
  power[0] = 1;
  for (int i = 1; i < n; i++) {
    power[i] = (power[i - 1] * B) % MOD;
  }
  preHash[0] = s[0] - 'a' + 1;
  for (int i = 1; i < n; i++) {
    preHash[i] = ((preHash[i - 1] * B) % MOD + (s[i] - 'a' + 1)) % MOD;
  }
  sufHash[n - 1] = s[n - 1] - 'a' + 1;
  for (int i = n - 2; i >= 0; i--) {
    sufHash[i] = ((sufHash[i + 1] * B) % MOD + (s[i] - 'a' + 1)) % MOD;
  }
  memset(dp, -1, sizeof dp);
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", solve(l - 1, r - 1));
  }
  return 0;
}
