题意：

给你两个序列 $a, b$ ，每次可以执行一个操作：将 $a [i]$ 与 $a [i + 1]$ 交换，且让交换后的 $a [i] + 1, a [i + 1] - 1$ ，问将 $a$ 变成 $b$ 的最小操作数。不能变成的话输出 $- 1$ 。

思路：

考虑将 $a_i=a_i+i,b_i=b_i+i$ ，如果有解的话那么这两个序列必须相同。证明不是很明白，待补。
之后就是裸题了，用线段树维护一下位置求出来就好了。

// Problem: E. String Reversal
// Contest: Codeforces - Educational Codeforces Round 96 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1430/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
int a[N],b[N];
map<int,vector<int>>v;
struct Node {int l,r;LL cnt,lazy;
}tr[N<<2];void pushdown(int u) {if(!tr[u].lazy) return;LL lazy=tr[u].lazy; tr[u].lazy=0;tr[L].lazy+=lazy; tr[R].lazy+=lazy;tr[L].cnt+=lazy; tr[R].cnt+=lazy;
}void build(int u,int l,int r) {tr[u]={l,r,1,0};if(l==r) {tr[u].cnt=l;return;}build(L,l,Mid); build(R,Mid+1,r);
} void change(int u,int l,int r,int c) {if(tr[u].l>=l&&tr[u].r<=r) {tr[u].cnt+=c;tr[u].lazy+=c;return;}pushdown(u);if(l<=Mid) change(L,l,r,c);if(r>Mid) change(R,l,r,c);
}int query(int u,int l,int r) {if(tr[u].l>=l&&tr[u].r<=r) return tr[u].cnt;int ans=0;pushdown(u);if(l<=Mid) ans+=query(L,l,r);if(r>Mid) ans+=query(R,l,r);return ans;
}LL solve() {LL ans=0;build(1,1,n);for(int i=1;i<=n;i++) {int pos=v[b[i]].back(); v[b[i]].pop_back();LL now=query(1,pos,pos);ans+=now-i;change(1,1,pos,1);}return ans;
}bool check() {map<int,int>mp1,mp2;for(int i=1;i<=n;i++) mp1[a[i]]++;for(int i=1;i<=n;i++) mp2[b[i]]++;for(int i=1;i<=n;i++) if(mp1[a[i]]!=mp2[a[i]]) return false;return true;
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);int _=1;while(_--) {cin>>n;for(int i=1;i<=n;i++) scanf("%d",&a[i]),a[i]+=i;for(int i=1;i<=n;i++) scanf("%d",&b[i]),b[i]+=i;for(int i=n;i>=1;i--) v[a[i]].pb(i);if(!check()) {puts("-1");return 0;}printf("%lld\n",solve());}return 0;
}
/**/