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 mod = 1e9 + 7;
const int maxm = 5e3 + 10;
int trans[maxm][30], fail[maxm], cnt[maxm], num[maxm], len[maxm], s[maxm];
int ans[maxm][maxm];
int n, p, last, sum;
char str[maxm];
int newnode(int l) {
  for (int i = 0; i < 26; i++) trans[p][i] = 0;
  cnt[p] = 0;
  num[p] = 0;
  len[p] = l;
  return p++;
}
void init() {
  sum = 0;
  p = 0;
  newnode(0);
  newnode(-1);
  last = 0;
  n = 0;
  s[n] = -1;
  fail[0] = 1;
}
int getfail(int x) {
  while (s[n - len[x] - 1] != s[n]) x = fail[x];
  return x;
}
void add(int c) {
  s[++n] = c;
  int cur = getfail(last);
  if (!trans[cur][c]) {
    int now = newnode(len[cur] + 2);
    fail[now] = trans[getfail(fail[cur])][c];
    trans[cur][c] = now;
    num[now] = num[fail[now]] + 1;
  }
  last = trans[cur][c];
  cnt[last]++;
  sum += num[last];
}
int main() {
  scanf("%s", str + 1);
  int length = strlen(str + 1);
  for (int i = 1; i <= length; i++) {
    init();
    for (int j = i; j <= length; j++) {
      add(str[j] - 'a');
      ans[i][j] = sum;
    }
  }
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", ans[l][r]);
  }
  return 0;
}
