/**************************************************************************************
	Palindrome tree. Useful structure to deal with palindromes in strings. O(N)
	This code counts number of palindrome substrings of the string.
	Based on problem 1750 from informatics.mccme.ru:
	http://informatics.mccme.ru/moodle/mod/statements/view.php?chapterid=1750
**************************************************************************************/

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <utility>
#include <cstring>
#include <cassert>
#include <cmath>
#include <stack>
#include <queue>

using namespace std;

const int MAXN = 105000;

struct node {
	int next[26];
	int len;
	int sufflink;
	int num;
};

int len;
char s[MAXN];
node tree[MAXN];
int num; 			// node 1 - root with len -1, node 2 - root with len 0
int suff; 			// max suffix palindrome
long long ans;

bool addLetter(int pos) {
	int cur = suff, curlen = 0;
	int let = s[pos] - 'a';

	while (true) {
		curlen = tree[cur].len;
		if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos])
			break;
		cur = tree[cur].sufflink;
	}
	if (tree[cur].next[let]) {
		suff = tree[cur].next[let];
		return false;
	}

	num++;
	suff = num;
	tree[num].len = tree[cur].len + 2;
	tree[cur].next[let] = num;

	if (tree[num].len == 1) {
		tree[num].sufflink = 2;
		tree[num].num = 1;
		return true;
	}

	while (true) {
		cur = tree[cur].sufflink;
		curlen = tree[cur].len;
		if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos]) {
			tree[num].sufflink = tree[cur].next[let];
			break;
		}
	}

	tree[num].num = 1 + tree[tree[num].sufflink].num;

	return true;
}

void initTree() {
	num = 2; suff = 2;
	tree[1].len = -1; tree[1].sufflink = 1;
	tree[2].len = 0; tree[2].sufflink = 1;
}

int main() {
	assert(freopen("pscpld.in", "r", stdin));
	assert(freopen("pscpld.out", "w", stdout));

	gets(s);
	len = strlen(s);

	initTree();

	for (int i = 0; i < len; i++) {
		addLetter(i);
		ans += tree[suff].num;
	}

	cout << ans << endl;

	return 0;
}