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 a[5005];
int dp[5005][5005];
int pp[5005][5005];
int main() {
  int i, j, n, m, k;
  cin >> a;
  n = strlen(a);
  for (i = 0; i < n; i++) pp[i][i] = 1;
  for (k = 2; k <= n; k++) {
    for (i = 0; i < n; i++) {
      if (i + k - 1 >= n) break;
      if (k == 2 && a[i] == a[i + k - 1])
        pp[i][i + k - 1] = 1;
      else {
        if (k > 2 && pp[i + 1][i + k - 1 - 1] && a[i] == a[i + k - 1])
          pp[i][i + k - 1] = 1;
      }
    }
  }
  for (i = 0; i < n; i++) dp[i][i] = 1;
  for (k = 2; k <= n; k++) {
    for (i = 0; i < n; i++) {
      if (i + k - 1 >= n) break;
      j = i + k - 1;
      dp[i][j] += dp[i][j - 1] + dp[i + 1][j];
      if (i + 1 <= j - 1) dp[i][j] -= dp[i + 1][j - 1];
      dp[i][j] += pp[i][j];
    }
  }
  cin >> m;
  for (i = 0; i < m; i++) {
    scanf("%d%d", &k, &j);
    printf("%d\n", dp[k - 1][j - 1]);
  }
  return 0;
}
