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 maxN = 5000 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
int d[maxN][maxN];
int pal[maxN][2];
int b[maxN];
int main() {
  ios_base::sync_with_stdio(false);
  string s;
  cin >> s;
  int n = ((int)(s).size());
  for (int i = 0; i < n; i++) {
    int p1, p2;
    p1 = p2 = i;
    while (p1 >= 0 && p2 < n && s[p1] == s[p2]) p1--, p2++, pal[i][0]++;
    p1 = i;
    p2 = i + 1;
    while (p1 >= 0 && p2 < n && s[p1] == s[p2]) p1--, p2++, pal[i][1]++;
  }
  for (int r = 0; r < n; r++) {
    for (int i = 0; i < n; i++) b[i] = 0;
    int cnt = 0;
    for (int l = r; l >= 0; l--) {
      int x = l - min(pal[l][0], r - l + 1);
      if (x >= 0) b[x]--;
      x = l - min(pal[l][1], r - l);
      if (x >= 0) b[x]--;
      cnt += 2 + b[l];
      d[l][r] = cnt + (l < r ? d[l + 1][r] : 0);
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << d[l][r] << endl;
  }
  return 0;
}
