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 std::min;
char str[5010], a[5010 << 1];
int n, r[5010 << 1], ispal[5010][5010];
inline void manacher() {
  for (int i = n; i >= 1; --i) a[i << 1] = str[i], a[(i << 1) - 1] = '#';
  a[0] = '$';
  int maxright = 0, mid = 0;
  for (int i = 1; i <= (n << 1); ++i) {
    if (i < maxright)
      r[i] = min(r[2 * mid - i], maxright - i + 1);
    else
      r[i] = 1;
    while (a[i - r[i]] == a[i + r[i]]) ++r[i];
    if (maxright < i + r[i] - 1) mid = i, maxright = i + r[i] - 1;
  }
  for (int i = 1; i <= (n << 1); ++i) {
    for (int j = 1; j <= r[i]; ++j) {
      int l = i - j + 1, r = i + j - 1;
      if (a[l] == '#' && a[r] == '#') ++l, --r;
      if (l > r) continue;
      ispal[l >> 1][r >> 1] = 1;
    }
  }
}
int ans[5010][5010];
int main() {
  scanf("%s", str + 1);
  n = strlen(str + 1);
  manacher();
  for (int r = 1; r <= n; ++r) {
    for (int l = 1; l <= r; ++l) {
      ispal[l][r] += ispal[l - 1][r];
    }
  }
  for (int l = 1; l <= n; ++l) {
    int res = 0;
    for (int r = l; r <= n; ++r) {
      res += ispal[r][r] - ispal[l - 1][r];
      ans[l][r] = res;
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", ans[l][r]);
  }
  return 0;
}
