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>
int a[5001][5001] = {0}, b[5001][5001] = {0};
int sum(int x, int y) {
  int i = 0, j = 0, s = 0;
  if (x > -1 && y > -1) {
    s = a[x][y];
    if (x > 0) {
      s = s + b[x - 1][y];
      if (y > 0) s = s - b[x - 1][y - 1];
    }
    if (y > 0) s = s + b[x][y - 1];
    return s;
  }
}
int main() {
  char s[5001];
  int l = 0, q = 0, i = 0, p = 0, j = 0;
  gets(s);
  l = strlen(s);
  for (i = 0; i < l; i++) {
    p = 0;
    while (i >= p && p <= (l - 1) / 2) {
      if (s[i - p] == s[i + p]) {
        a[i - p][i + p]++;
        p++;
      } else
        break;
    }
    q = 0;
    while (i >= q && i <= (2 * l - 3) / 2) {
      if (s[i - q] == s[i + 1 + q]) {
        a[i - q][i + 1 + q]++;
        q++;
      } else
        break;
    }
  }
  for (i = 0; i < l; i++) {
    for (j = 0; j < l; j++) {
      b[i][j] = sum(i, j);
    }
  }
  int t, count;
  scanf("%d", &t);
  for (count = 0; count < t; count++) {
    int n = 0, m = 0, z = 0;
    scanf("%d%d", &n, &m);
    n--;
    m--;
    z = b[m][m];
    if (n > 0) z = z + b[n - 1][n - 1] - b[m][n - 1] - b[n - 1][m];
    printf("%d\n", z);
  }
  return 0;
}
