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, sz;
string seq;
int dp[5050][5050];
int dr[5050][5050];
bool pal[5050][5050];
int main() {
  cin >> seq;
  sz = seq.size();
  for (int i = 0; i < sz; i++) pal[i][i] = 1;
  for (int p = 1; p < sz; p++) {
    for (int i = 0; i < sz; i++) {
      int j = i + p;
      if (j >= sz) break;
      if (pal[i + 1][j - 1] && seq[i] == seq[j])
        pal[i][j] = 1;
      else if (i + 1 == j && seq[i] == seq[j])
        pal[i][j] = 1;
    }
  }
  memset(dp, 0, sizeof dp);
  for (int i = 0; i < sz; i++) {
    dp[i][i] = pal[i][i];
    for (int j = i + 1; j < sz; j++) dp[i][j] = dp[i][j - 1] + pal[i][j];
  }
  for (int j = sz - 1; j >= 0; j--) {
    dr[j][j] = dp[j][j];
    for (int i = j - 1; i >= 0; i--) {
      dr[i][j] = dr[i + 1][j] + dp[i][j];
    }
  }
  scanf("%d", &n);
  while (n--) {
    int l, r;
    scanf("%d %d", &l, &r);
    l--;
    r--;
    printf("%d\n", dr[l][r]);
  }
  return 0;
}
