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 int N = 5000 + 1;
char text[N];
int sum[N][N];
int main() {
  scanf("%s", text);
  int n = strlen(text);
  for (int middle = 0; middle <= (n - 1) << 1; ++middle) {
    int left = middle >> 1;
    int right = middle - left;
    while (left >= 0 && right < n) {
      if (text[left] != text[right]) {
        break;
      }
      sum[left][right]++;
      left--;
      right++;
    }
  }
  for (int i = n - 1; i >= 0; --i) {
    for (int j = i; j < n; ++j) {
      long long result = sum[i][j];
      if (j - i + 1 > 1) {
        result += sum[i + 1][j];
        result += sum[i][j - 1];
      }
      if (j - i + 1 > 2) {
        result -= sum[i + 1][j - 1];
      }
      sum[i][j] = result;
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    r--;
    printf("%d\n", sum[l][r]);
  }
  return 0;
}
