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 long double PI = acos(-1);
const int MOD = 1000000007;
const int FMOD = 998244353;
long long int dp[5005][5005];
char s[1000006];
void solve() {
  scanf("%s", s + 1);
  int n;
  for (n = 1; s[n]; n++)
    ;
  n--;
  for (int i = 1; i <= n; i++) {
    dp[i][i] = 1;
    if (i > 1) {
      if (s[i] == s[i - 1]) {
        dp[i - 1][i]++;
      }
    }
  }
  for (int i = n - 2; i >= 1; i--) {
    for (int j = i + 2; j <= n; j++) {
      if (s[i] == s[j]) {
        dp[i][j] = dp[i + 1][j - 1];
      }
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    for (int j = i + 1; j <= n; j++) {
      dp[i][j] += dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1];
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%lld\n", dp[l][r]);
  }
}
signed main() { solve(); }
