// Laufzeit: O(n*log(n))
ll merge(vector<ll>& v, vector<ll>& left, vector<ll>& right) {
	int a = 0, b = 0, i = 0;
	ll inv = 0;
	while (a < (int)left.size() && b < (int)right.size()) {
		if (left[a] < right[b]) v[i++] = left[a++];
		else {
			inv += left.size() - a;
			v[i++] = right[b++];
		}
	}
	while (a < (int)left.size()) v[i++] = left[a++];
	while (b < (int)right.size()) v[i++] = right[b++];
	return inv;
}

ll mergeSort(vector<ll> &v) { // Sortiert v und gibt Inversionszahl zurück.
	int n = v.size();
	vector<ll> left(n / 2), right((n + 1) / 2);
	for (int i = 0; i < n / 2; i++) left[i] = v[i];
	for (int i = n / 2; i < n; i++) right[i - n / 2] = v[i];

	ll result = 0;
	if (left.size() > 1) result += mergeSort(left);
	if (right.size() > 1) result += mergeSort(right);
	return result + merge(v, left, right);
}