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 n, l, r;
int dp[5001][5001];
bool p[5001][5001];
char buf[10000];
int max(int a, int b) { return a > b ? a : b; }
void solve(char *buf) {
  int i, j, len;
  len = strlen(buf + 1);
  memset(p, 0, sizeof(p));
  for (i = 1; i <= len; i++) p[i][i] = true;
  for (j = 1; j < len; j++)
    for (i = 1; i + j <= len; i++)
      if (buf[i] == buf[i + j])
        if (i + 1 < i + j - 1)
          p[i][i + j] = p[i + 1][i + j - 1];
        else
          p[i][i + j] = true;
  memset(dp, 0, sizeof(dp));
  for (i = 1; i <= len; i++) dp[i][i] = 1;
  for (j = 1; j < len; j++)
    for (i = 1; i + j <= len; i++) {
      dp[i][i + j] = dp[i + 1][i + j] + dp[i][i + j - 1] - dp[i + 1][i + j - 1];
      if (p[i][i + j]) dp[i][i + j] += 1;
    }
}
int main() {
  while (scanf("\n%s", buf + 1) != EOF) {
    solve(buf);
    scanf("%d", &n);
    while (n-- > 0) {
      scanf("%d%d", &l, &r);
      printf("%d\n", dp[l][r], p[l][r]);
    }
  }
  return 0;
}
