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 N, M;
int dpp[5005][5005];
int dp[5005][5005];
string A;
bool Dp(int x, int y) {
  if (x == y) return 1;
  if (x + 1 == y) return A[x] == A[y];
  int &ret = dp[x][y];
  if (ret != -1) return ret;
  return ret = Dp(x + 1, y - 1) && A[x] == A[y];
}
int Dp2(int x, int y) {
  if (x > y) return 0;
  int &ret = dpp[x][y];
  if (ret != -1) return ret;
  return ret = Dp2(x, y - 1) - Dp2(x + 1, y - 1) + Dp2(x + 1, y) + Dp(x, y);
}
int main() {
  int x, y;
  cin >> A >> N;
  memset(dp, -1, sizeof dp);
  memset(dpp, -1, sizeof dpp);
  for (int i = 1; i <= N; i++) {
    scanf("%d %d", &x, &y);
    printf("%d\n", Dp2(x - 1, y - 1));
  }
  return 0;
}
