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;
char ch[5005];
int n;
bool can[5005][5005];
int dp[5005][5005];
int s[5005];
int main() {
  cin >> ch;
  n = strlen(ch);
  for (int i = 0; i < n; i++) {
    can[i][i] = 1;
    for (int j = 1; j <= i && i + j < n; j++)
      can[i - j][i + j] = can[i - j + 1][i + j - 1] && (ch[i - j] == ch[i + j]);
    if (i < n - 1) {
      can[i][i + 1] = (ch[i] == ch[i + 1]);
      for (int j = 1; j <= i && i + j + 1 < n; j++)
        can[i - j][i + j + 1] =
            can[i - j + 1][i + j] && (ch[i - j] == ch[i + j + 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) s[i] += can[i][j];
  }
  for (int i = n - 1; i >= 0; i--) {
    dp[i][i] = 1;
    for (int j = i - 1; j >= 0; j--) {
      dp[j][i] = dp[j + 1][i] + s[j];
      s[j] -= can[j][i];
    }
  }
  int m;
  cin >> m;
  while (m--) {
    int a, b;
    cin >> a >> b;
    cout << dp[a - 1][b - 1] << endl;
  }
}
