#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
// #include <ext/pb_ds/tree_policy.hpp>
// #include <ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;
// template <class T> using Tree = tree<T, null_type, less<T>,
// rb_tree_tag, tree_order_statistics_node_update>;
const int maxn = 100002;
#define mp make_pair
#define pb push_back
#define f first
#define s second
const int mod = 1e9+7;
typedef pair<int, int> pii;
int n;
// for each shoe, we need the closest shoe to the right that is the opposite pair to it
int r[2*maxn];
vector<int> eachsize[2*maxn]; //eachsize[i] contains the positions of the shoes of size i-n in order
int sizepos[2*maxn]; //sizepos[i] tells us how many shoes of size i-n we've seen
int fentree[2*maxn];
/*
We need to figure out using some PURS thing how to get the number of elements between i and r[i] that have already been taken care of
we can do this by just making a BIT where initially everything is 1, and then if the element i has been used, we make BIT[i] = 0
*/
int lsb(int a)
{
return a & -a;
}
void change(int a, int c)
{
while (a <= 2*n)
{
fentree[a] += c;
a += lsb(a);
}
}
ll sum(int a)
{
ll s = 0;
while (a > 0)
{
s += fentree[a];
a -= lsb(a);
}
return s;
}
ll count_swaps(vector<int> shoes)
{
n = shoes.size()/2;
shoes.insert(shoes.begin(), 0);
for (int i=1; i<=2*n; i++)
{
eachsize[shoes[i]+n].pb(i);
}
for (int i=1; i<=2*n; i++)
{
if (sizepos[shoes[i]+n] < (int) eachsize[-shoes[i]+n].size())
{
int temp = eachsize[-shoes[i]+n][sizepos[shoes[i]+n]];
if (temp > i)
{
r[i] = temp;
}
}
sizepos[shoes[i]+n]++;
}
ll ans = 0;
for (int i=1; i<=2*n; i++)
{
change(i, 1);
}
for (int i=1; i<=2*n; i++)
{
if (r[i] != 0)
{
if (r[i] != i+1)
{
ans += sum(r[i]-1) - sum(i);
//cerr << sum(r[i]-1) << " " << sum(i) << endl;
}
if (shoes[i] > 0)
{
ans++;
}
change(r[i], -1);
}
}
return ans;
}
