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 str[5000 + 10];
int dp[5000 + 1][5000 + 1] = {{0}};
short isp[5000 + 1][5000 + 1] = {{0}};
int main() {
  while ((scanf("%s", str + 1)) != EOF) {
    memset(dp, 0, sizeof(dp));
    memset(isp, 0, sizeof(isp));
    int len = strlen(str + 1);
    for (int i = 1; i <= len; i++) isp[i][i] = isp[i][i - 1] = dp[i][i] = 1;
    for (int r = 2; r <= len; r++) {
      for (int i = 1; i < len; i++) {
        int j = i + r - 1;
        isp[i][j] = (isp[i + 1][j - 1] && str[i] == str[j]);
        dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1] + isp[i][j];
      }
    }
    int n;
    cin >> n;
    int a, b;
    for (int i = 1; i <= n; i++) {
      scanf("%d%d", &a, &b);
      printf("%d\n", dp[a][b]);
    }
  }
  return 0;
}
