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/*
 *  Copyright (C) 2018  Paweł Widera
 *
 *  This program is free software; you can redistribute it and/or modify
 *  it under the terms of the GNU General Public License as published by
 *  the Free Software Foundation; either version 3 of the License, or
 *  (at your option) any later version.
 *
 *  This program is distributed in the hope that it will be useful,
 *  but WITHOUT ANY WARRANTY; without even the implied warranty of
 *  MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 *  GNU General Public License for more details:
 *  http://www.gnu.org/licenses/gpl.html
 */
#include <iostream>
#include <string>
#include <vector>
#include <unordered_set>
using namespace std;

#define DBG(key, value) cerr << key << " " << value << endl;

const unordered_set<char> vowels = {'a', 'e', 'i', 'o', 'u', 'y'};


int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	string text;
	cin >> text;

	long long int n = text.size();
	if (n < 3) {
		cout << 0 << endl;
		return 0;
	}

	vector<pair<long long int,long long int>> ranges;
	int vowel = 0;
	int consonant = 0;

	// find all difficult ranges
	for (int i = 0; i < n; ++i) {
		if (vowels.count(text[i])) {
			++vowel;
			consonant = 0;
		} else {
			++consonant;
			vowel = 0;
		}
		if (consonant >= 3 || vowel >= 3) {
			int length = max(consonant, vowel);
			int start = i - length + 1;

			if (!ranges.empty() && ranges.back().first == start) {
				ranges.pop_back();
			}
			ranges.emplace_back(start, i + 1);
		}
	}

	int last = 0;

	// count all fragments around each range
	long long int counter = 0;
	for (auto range : ranges) {
		auto length = range.second - range.first;
		// fragments inside the range
		counter += (length - 2) * (length - 1) / 2 - 1;
		// fragments containing the range
		counter += (range.first - last + 1) * (n - range.second + 1);
		// fragment that begin/end within the range
		if (length > 3) {
			counter += (range.first - last) * (length - 3);
			counter += (n - range.second) * (length - 3);
		}
		// remember current end to avoid counting overlaps
		last = range.second - 2;
	}

	cout << counter << endl;
	return 0;
}