Queries for Number of Palindromes(求任意子列的回文数)

H. Queries for Number of Palindromes

time limit per test

 5 seconds

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

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.

Sample test(s)

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».

代码：

#include<stdio.h>
#include<string.h>
int dp[5010][5010],and1[5010][5010];
int main()
{
    char s[5010];
    int q,l,r,sta,end,len;
    int i;
    gets(s);
    len=strlen(s);
    for(i=0; i<len; i++)
    {
        dp[i][i]=and1[i][i]=1;
        and1[i+1][i]=1;
    }
    for(i=2;i<=len;i++)
        for(sta=0;sta<len;sta++)
        {
            end=sta+i-1;
            if(and1[sta+1][end-1]&&s[sta]==s[end])
                and1[sta][end]=1;
            dp[sta][end]=dp[sta+1][end]+dp[sta][end-1]-dp[sta+1][end-1]+and1[sta][end];
        }
    scanf("%d",&q);
    while(q--)
    {
        scanf("%d %d",&l,&r);
        printf("%d\n",dp[l-1][r-1]);
    }
    return 0;
}