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 l_i, r_i (1 ≤ l_i ≤ r_i ≤ |s|). The answer to the query is the number of substrings of string s[l_i... r_i], which are palindromes.

String s[l... r] = s_ls_l + 1... s_r (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 l_i, r_i (1 ≤ l_i ≤ r_i ≤ |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

caaaba
5
1 1
1 4
2 3
4 6
4 5

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».

题意：求一个区间内回文串的数量

思路：要利用一个判断l~r区间的回文数组来判断这个区间是否是回文串是则为1不是则为0 然后定义一个dp数组来记录l~r区间内回文串的数量

代码：

#include<cstdio>

#include<iostream>

#include<cstring>

#include<algorithm>

#include<queue>

#include<stack>

#include<set>

#include<vector>

#include<map>

#include<cmath>

const int maxn=1e5+;

typedef long long ll;

using namespace std;

int hw[][];

int dp[][];

char str[];

int main()

{

   scanf("%s",str+);

   int len=strlen(str+);

   for(int t=;t<=len;t++)

   {

       hw[t][t]=;

       hw[t][t-]=;

       dp[t][t]=;

   }

   for(int d=;d<=len;d++)

   {

       for(int l=;l+d<=len;l++)

       {

           int r=l+d;

           if(str[l]==str[r]&&hw[l+][r-])

           {

               hw[l][r]=;

        }

    }

   }

   for(int d=;d<=len;d++)

   {

       for(int l=;l+d<=len;l++)

       {

           int r=l+d;

           dp[l][r]=dp[l+][r]+dp[l][r-]-dp[l+][r-]+hw[l][r];

    }

   }

   int q;

   cin>>q;

   int l,r;

   while(q--)

   {

       scanf("%d%d",&l,&r);

       printf("%d\n",dp[l][r]);

   }

   return ;

}