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;
class Q {
 public:
  int a, b, ind;
  bool operator<(const Q &q) const { return b < q.b; }
};
Q qs[1000100];
int ans[1000100];
bool eh[5010][5010];
int bit[5010];
char s[5010];
void add(int p, int v) {
  for (; p < 5010; p += (p & (-p))) bit[p] += v;
}
int get(int p) {
  int r = 0;
  for (; p; p -= (p & (-p))) r += bit[p];
  return r;
}
int main() {
  scanf("%s", s + 1);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d %d", &qs[i].a, &qs[i].b);
    qs[i].ind = i;
  }
  sort(qs, qs + q);
  int n = strlen(s + 1);
  for (int l = 1; l < n + 1; l++)
    for (int i = 1; i < n + 2 - l; i++) {
      int j = i + l - 1;
      if (i == j)
        eh[i][j] = true;
      else if (s[i] == s[j] && (i == j - 1 || eh[i + 1][j - 1]))
        eh[i][j] = true;
    }
  int p = 1;
  for (int i = 0; i < q; i++) {
    while (p <= qs[i].b) {
      for (int j = 1; j < p + 1; j++) {
        if (eh[j][p]) {
          add(1, 1), add(j + 1, -1);
        }
      }
      p++;
    }
    ans[qs[i].ind] = get(qs[i].a);
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  return 0;
}
