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;
using ll = long long;
const int N = 5005;
int fw[N][N];
int que(int x, int y) {
  int ret = 0;
  for (; x; x -= x & (-x)) {
    int y1 = y;
    for (; y1; y1 -= y1 & (-y1)) {
      ret += fw[x][y1];
    }
  }
  return ret;
}
void update(int x, int y) {
  for (; x < N; x += x & (-x)) {
    int y1 = y;
    for (; y1 < N; y1 += y1 & (-y1)) {
      fw[x][y1]++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  int n = s.length();
  vector<vector<bool>> dp(n, vector<bool>(n, 0));
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  for (int len = 2; len <= n; len++) {
    for (int i = 0; i + len <= n; i++) {
      int j = i + len - 1;
      if (s[i] != s[j]) continue;
      if (len == 2)
        dp[i][j] = true;
      else
        dp[i][j] = dp[i + 1][j - 1];
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (dp[i][j]) update(i + 1, j + 1);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    int ret = que(r, r);
    if (l != 1) {
      ret -= que(l - 1, r);
      ret -= que(r, l - 1);
      ret += que(l - 1, l - 1);
    }
    cout << ret << "\n";
  }
  return 0;
}
