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;
int n;
int dp[5005][5005];
bool isPalin[5005][5005];
int sum[5005][5005];
string s;
int palin(int left, int right) {
  int &res = sum[left][right];
  if (res == -1) {
    res = 0;
    int len = right - left + 1;
    if (len <= 2) {
      if (s[left] == s[right])
        res = 1;
      else
        res = 0;
    } else {
      if (s[left] == s[right])
        res = palin(left + 1, right - 1);
      else
        res = 0;
    }
  }
  return res;
}
int findSum(int left, int right) {
  if (left == 0) return sum[right][right];
  return sum[right][right] - sum[left - 1][right];
}
void compute() {
  memset(sum, -1, sizeof(sum));
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      isPalin[i][j] = palin(i, j);
    }
  }
  memset(sum, 0, sizeof(sum));
  for (int j = 0; j < n; j++) {
    int cnt = 0;
    if (isPalin[0][j]) cnt = 1;
    sum[0][j] = cnt;
    for (int i = 1; i <= j; i++) {
      if (isPalin[i][j]) cnt++;
      sum[i][j] = cnt;
    }
  }
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      dp[i][j] = dp[i][j - 1] + findSum(i, j);
    }
  }
}
int main() {
  cin >> s;
  n = s.size();
  compute();
  int q;
  cin >> q;
  while (q--) {
    int left, right;
    cin >> left >> right;
    left--;
    right--;
    cout << dp[left][right] << endl;
  }
  return 0;
}
