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>
const int maxn = 5003;
bool flag[maxn][maxn];
int sum[maxn][maxn];
char s[maxn];
int main() {
  while (scanf("%s", s) == 1) {
    int n = strlen(s);
    for (int i = 0; i < n; i++) flag[i][i] = 1;
    for (int i = 0; i < n - 1; i++)
      if (s[i] == s[i + 1]) flag[i][i + 1] = 1;
    for (int d = 2; d < n; d++)
      for (int i = 0; i < n; i++)
        if (i + d < n)
          if (flag[i + 1][i + d - 1] && s[i] == s[i + d]) flag[i][i + d] = 1;
    for (int d = 0; d < n; d++)
      for (int i = 0; i < n; i++)
        if (i + d < n) {
          int j = i + d;
          if (j)
            sum[i][j] = sum[i][j - 1] + sum[i + 1][j] - sum[i + 1][j - 1];
          else
            sum[i][j] = 0;
          sum[i][j] += flag[i][j];
        }
    int q;
    scanf("%d", &q);
    int l, r;
    while (q--) {
      scanf("%d%d", &l, &r);
      printf("%d\n", sum[l - 1][r - 1]);
    }
  }
  return 0;
}
