#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
#define SS ios_base::sync_with_stdio(0);cin.tie(nullptr);cout.tie(nullptr);
#pragma optimize("g", on)
#pragma GCC optimize("03")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,avx2,mmx,fma,avx,tune=native")
#define int long long
#define all(x) x.begin(),x.end()
#define F first
#define S second
using namespace std;
// using namespace __gnu_pbds;
// #define ordered_set tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update>
const int LG = 21,N = 2e5+1,inf = 1e18,MOD = 998244353;
const double eps = 1e-9;
int T;
int f[N];
int n;
void add(int i,int x){
while(i<=n){
f[i]+=x;
i+=i&-i;
}
}
int get(int i){
int res=0;
while(i){
res+=f[i];
i-=i&-i;
}
return res;
}
int range(int l,int r){
return get(r)-get(l);
}
void solve() {
cin>>n;
int h[n+1];
int id[n+1];
for(int i=1;i<=n;i++){
cin>>h[i];
id[h[i]]=i;
}
int cost[n+2][n+2]={},pref[n+2][n+2]={};
for(int r=1;r<=n;r++){
for(int i=1;i<=r;i++) pref[r][id[i]]++;
for(int i=1;i<=n;i++){
pref[r][i]+=pref[r][i-1];
}
for(int l=r;l>=1;l--){
cost[l][r]=cost[l+1][r]+(pref[r][n]-pref[r][id[l]])-get(id[l]);
if(r<id[l]) cost[l][r]+=id[l]-r;
add(id[l],1);
}
for(int i=1;i<=n;i++) f[i]=0;
}
vector<int> dp(n+1,inf);
dp[0]=0;
for(int r=1;r<=n;r++){
for(int l=r;l>=1;l--){
dp[r]=min(dp[r],dp[l-1]+cost[l][r]);
}
}
cout<<dp[n];
return;
}
signed main() {
// freopen("deleg.in","r",stdin);
// freopen("deleg.out","w",stdout);
SS
int t = 1;
if (T) {
cin >> t;
}
while (t--) {
solve();
}
}
/*
1,2,3,...,n-2,n-1,n
good array is reverse of subsequence of subarray
because the condition a[i]-a[i+1]<=1 must hold
PROOF
{
Suppose on the position i there is number x
the next number which we can pick is x-1 or y>x
when we take number which is greater than x, x-1 number must be taken on prefix
Otherwise y>x will only decrease by 1 and the minimum number on suffix will x+1, and x+1-x-1=2 condition doesnt hold
}
dp[i][j] -> prefix i, last number on reversed array
1<=j<=i dp[i] = dp[j-1] + op[j]
*/
