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_MAX = 5050;
char s[N_MAX];
int memo[N_MAX][N_MAX];
bitset<N_MAX> pal[N_MAX];
bitset<N_MAX> vis[N_MAX];
bool isPali(int i, int j) {
  if (i > j) return 0;
  if (vis[i][j]) return pal[i][j];
  vis[i][j] = 1;
  pal[i][j] = (s[i] == s[j]);
  if (j - i < 2) return pal[i][j];
  return pal[i][j] = (pal[i][j] && isPali(i + 1, j - 1));
}
int solve(int i, int j) {
  if (i > j) swap(i, j);
  if (memo[i][j] != -1) return memo[i][j];
  if (i == j) return memo[i][j] = 1;
  if (i + 1 == j) return memo[i][j] = 2 + isPali(i, j);
  return memo[i][j] = solve(i + 1, j) + solve(i, j - 1) - solve(i + 1, j - 1) +
                      isPali(i, j);
}
int main() {
  memset(memo, -1, sizeof(memo));
  scanf("%s", s);
  int q;
  scanf("%d", &q);
  while (q--) {
    int i, j;
    scanf("%d%d", &i, &j);
    printf("%d\n", solve(i - 1, j - 1));
  }
}
