Codeforces 245H Queries for Number of Palindromes （區間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 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».

 

題意：給你一個字符串，q次查詢，問你這個區間的迴文串個數多少？

解題思路：

求解區間問題，如果不超過存儲範圍就可以用區間dp。先預處理每個區間是否是迴文串。然後區間dp尋找回文串個數

如果長度爲2的區間（l==1) 那麼很明顯 最多隻有三種迴文個數

否則的話 那麼我們就需要根據容斥原理計算一下i+1到 j-1範圍內迴文個數  加上當前這整個區間是否是迴文串

最後查詢 0（1）輸出

#include<stdio.h>
#include<String.h>
#include<algorithm>
using namespace std;
const int maxn=5e3+10;
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
char ch[maxn];
int n,m,dp[maxn][maxn];
bool Is[maxn][maxn];
int main(){
	int i,j,l;
	scanf("%s",ch);
	int len=strlen(ch);
	mem(Is,0);mem(dp,0);
	for(i=0;i<len;i++) Is[i][i]=true;
	for(l=1;l<len;l++){
		for(i=0;i<len-l;i++){
			j=i+l;
			if(ch[i]==ch[j]){
				if(l==1) Is[i][j]=true;
				else{
					if(Is[i+1][j-1]) Is[i][j]=true;
				}
			}
		}
	}
	for(i=0;i<len;i++) dp[i][i]=1;
	for(l=1;l<len;l++){
		for(i=0;i<len-l;i++){
			j=i+l;
			if(l==1) dp[i][j]=dp[i][i]+dp[j][j]+Is[i][j];
			else dp[i][j]=dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+Is[i][j];
		}
	}
	scanf("%d",&m);
	while(m--){
		int l,r;
		scanf("%d%d",&l,&r);
		printf("%d\n",dp[l-1][r-1]);
	}
	return 0;
}