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 dp[5005][5005], W[5005][5005], q, x, y;
bool jyz[5005][5005];
string s;
int main() {
  ;
  ;
  ;
  ;
  ;
  cin >> s;
  int n = s.size();
  for (int i = 0; i <= n; i++) jyz[i][i] = 1;
  for (int i = 0; i <= n - 1; i++)
    if (s[i] == s[i + 1]) jyz[i][i + 1] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 0; j + i < n; j++)
      if (jyz[j + 1][j + i - 1] == 1 && s[j] == s[j + i]) jyz[j][j + i] = 1;
  for (int i = 0; i <= n; i++)
    for (int j = i; j >= 0; j--)
      if (jyz[j][i] == 1)
        W[j][i] = W[j + 1][i] + 1;
      else
        W[j][i] = W[j + 1][i];
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j < n - i; j++)
      dp[j][i + j] = dp[j][i + j - 1] + W[j][i + j];
  for (cin >> q; q; q--)
    scanf("%d%d", &x, &y), printf("%d\n", dp[x - 1][y - 1]);
  return 0;
}
