Problem: Training is indispensable for achieving good results at ICPC. Rabbit wants to win at ICPC, so he decided to practice today as well.

Today's practice is to improve the ability to read messages hidden from sentences by searching for palindromes in strings. There may be many palindromes, so I'd like to count the number while searching.

Given two strings S and T, we want to find the number of pairs of integers (i, j, k, l) that satisfy the following:

* 1 ≤ i ≤ j ≤ (length of S).
* 1 ≤ k ≤ l ≤ (length of T).
* The substrings extracted from the i-th to j-th characters of S are the same as the substrings extracted from the k-th to l-th characters of T, and these are palindromes (even when read from the left). It is a character string that is the same even when read from the right).



Input


S
T


Both the strings S and T have a length of 1 or more and 50,000 or less, and consist of uppercase letters.

Output

Output the number of pairs of integers (i, j, k, l) that satisfy the condition on one line.

Examples

Input

ICPC
CPCPC


Output

10


Input

BABBAB
ABBA


Output

14


Input

MYON
USAGI


Output

0

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 99999999999999999
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;


#define NUM 100005

enum Type{
	S,
	T,
};

struct Node{
	    int parent_id,children[28],suffix_link,length,count;
};

int num_nodes,root,last_node;
ll dp[2][NUM];
char base_ch;
Node nodes[NUM];
string input_S,input_T,line;


void add(int loc){

	char tmp_ch = line[loc];

	int tmp_node = last_node;

	while(true){
		if(loc-1-nodes[tmp_node].length >= 0 && line[loc-1-nodes[tmp_node].length] == tmp_ch){
			break;
		}
		tmp_node = nodes[tmp_node].suffix_link;
	}

	if(nodes[tmp_node].children[tmp_ch-base_ch] > 0){

		last_node = nodes[tmp_node].children[tmp_ch-base_ch];
		nodes[last_node].count++;
		return;
	}

	last_node = num_nodes++;
	nodes[tmp_node].children[tmp_ch-base_ch] = last_node;

	nodes[last_node].length = nodes[tmp_node].length+2;
	nodes[last_node].parent_id = tmp_node;
	nodes[last_node].count = 1;

	if(nodes[last_node].length == 1){

		nodes[last_node].suffix_link = root+1;

	}else{

		while(true){
			tmp_node = nodes[tmp_node].suffix_link;
			if(loc-1-nodes[tmp_node].length >= 0 && line[loc-1-nodes[tmp_node].length] == tmp_ch){

				nodes[last_node].suffix_link = nodes[tmp_node].children[tmp_ch-base_ch];
				break;
			}
		}
	}
}

int get_index(){

	return last_node;
}

void init(string arg_line){

	line = arg_line;
	num_nodes = 0;

	base_ch = '?';

	root = 0;
	nodes[root].suffix_link = -1;

	for(int i = 0; i < NUM; i++){

		nodes[i].parent_id = -1;
		nodes[i].count = 0;

		for(int k = 0; k < 28; k++){
			nodes[i].children[k] = -1;
		}
	}

	num_nodes = 2;
	last_node = root;

	nodes[root].length = -1;
	nodes[root+1].length = 0;

	nodes[root].suffix_link = root;
	nodes[root+1].suffix_link = root;

	nodes[root].parent_id = -1;
	nodes[root+1].parent_id = root;
}

int main(){

	cin >> input_S >> input_T;

	string LINE = input_S + "?@" + input_T;

	init(LINE);

	int tmp_index;

	for(int i = 0; i < 2; i++){
		for(int k = 0; k < NUM; k++){

			dp[i][k] = 0;
		}
	}

	for(int i = 0; i < LINE.length(); i++){

		add(i);
		tmp_index = get_index();

		if(i < input_S.length()){

			dp[S][tmp_index]++;

		}else if(i >= input_S.length()+2){

			dp[T][tmp_index]++;
		}
	}

	ll ans = 0;

	for(int i = num_nodes-1; i >= 2; i--){

		ans += dp[S][i]*dp[T][i];
		dp[S][nodes[i].suffix_link] += dp[S][i];
		dp[T][nodes[i].suffix_link] += dp[T][i];
	}

	printf("%lld\n",ans);

    return 0;
}