Queries for Number of Palindromes（区间dp）

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

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<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>

const int maxn=5e3+5;
typedef long long ll;
using namespace std;
char str[maxn];

int a[maxn][maxn],dp[maxn][maxn];

int main()
{
    scanf("%s",str+1);
    int len=strlen(str+1);
    //cout<<len<<endl;
    for(int t=1;t<=len;t++)
    {
        a[t][t]=a[t][t-1]=dp[t][t]=1;
    }
    for(int t=2;t<=len;t++)
    {
        for(int j=1;j+t-1<=len;j++)
        {
            if(str[j]==str[j+t-1]&&a[j+1][j+t-2]%2==1)
            a[j][j+t-1]=1;
            else
            {
                a[j][j+t-1]=0;
            }
        }
    }
    for(int t=2;t<=len;t++)
    {
        for(int j=1;j+t-1<=len;j++)
        {
            dp[j][j+t-1]=dp[j][j+t-2]+dp[j+1][j+t-1]-dp[j+1][j+t-2]+a[j][j+t-1];
        }
    }
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        printf("%d
",dp[l][r]);
    }
    
    return 0;
}