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 maxn = 5005;
string s;
int q, n, i, j, a, b, l;
int dp[maxn][maxn], fix[maxn][maxn], A[maxn];
int main() {
  cin >> s;
  n = s.size();
  for (i = 1; i <= n; i++) fix[i][i] = 1;
  for (i = 1; i <= n - 1; i++) fix[i][i + 1] = (s[i] == s[i - 1]);
  for (l = 2; l <= n - 1; l++)
    for (j = 1; j + l <= n; j++) {
      a = j;
      b = j + l;
      fix[a][b] = (fix[a + 1][b - 1] && s[a - 1] == s[b - 1]);
    }
  for (i = 1; i <= n; i++) dp[i][i] = 1;
  for (i = 1; i <= n - 1; i++) dp[i][i + 1] = 2 + fix[i][i + 1];
  for (i = 1; i <= n - 1; i++) A[i] = fix[i][i + 1] + fix[i][i];
  for (l = 2; l <= n - 1; l++) {
    for (j = 1; j + l <= n; j++) A[j] = A[j] + fix[j][j + l];
    for (j = 1; j + l <= n; j++) {
      a = j;
      b = j + l;
      dp[a][b] = dp[a + 1][b] + A[a];
    }
  }
  cin >> q;
  while (q--) {
    cin >> a >> b;
    cout << dp[a][b] << endl;
  }
}
