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 MX = int(5e3) + 10;
char str[MX];
int rad[MX][2];
int l[MX][MX], r[MX][MX];
int n;
int init() {
  for (int i = 1; i <= n; i++) {
    rad[i][0] = 1;
    for (int j = 1; j <= min(i - 1, n - i); j++) {
      if (str[i - j] == str[i + j]) {
        rad[i][0]++;
      } else {
        break;
      }
    }
    for (int j = 1; j <= min(i, n - i); j++) {
      if (str[i + 1 - j] == str[i + j]) {
        rad[i][1]++;
      } else {
        break;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    l[i][i] = rad[i][0] + rad[i][1] + l[i - 1][i - 1];
    for (int j = i + 1; j <= n; j++) {
      l[i][j] = l[i][j - 1];
      l[i][j] += max(0, i - (j - rad[j][0]));
      l[i][j] += max(0, i - (j - rad[j][1]));
    }
  }
  for (int i = n; i > 0; i--) {
    r[i][i] = rad[i][0] + rad[i][1] + r[i + 1][i + 1];
    for (int j = i - 1; j > 0; j--) {
      r[i][j] = r[i][j + 1];
      r[i][j] += max(0, j + rad[j][0] - i);
      r[i][j] += max(0, j + rad[j][1] - i + 1);
    }
  }
  return r[1][1];
}
int main() {
  scanf("%s", str + 1);
  n = strlen(str + 1);
  int sum = init();
  int q;
  scanf("%d", &q);
  int x, y;
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &x, &y);
    x--;
    y++;
    printf("%d\n", min(sum - l[x][((x + y) >> 1) - 1] - r[y][((x + y) >> 1)],
                       sum - l[x][((x + y) >> 1)] - r[y][((x + y) >> 1) + 1]));
  }
}
