Educational Codeforces Round 78 (Rated for Div. 2) C. Berry Jam

Karlsson has recently discovered a huge stock of berry jam jars in the basement of the house. More specifically, there were 2n2n jars of strawberry and blueberry jam.

All the 2n2n jars are arranged in a row. The stairs to the basement are exactly in the middle of that row. So when Karlsson enters the basement, he sees exactly nn jars to his left and nn jars to his right.

For example, the basement might look like this:

Being the starightforward man he is, he immediately starts eating the jam. In one minute he chooses to empty either the first non-empty jar to his left or the first non-empty jar to his right.

Finally, Karlsson decided that at the end the amount of full strawberry and blueberry jam jars should become the same.

For example, this might be the result:

He has eaten 11 jar to his left and then 55 jars to his right. There remained exactly 33 full jars of both strawberry and blueberry jam.

Jars are numbered from 11 to 2n2n from left to right, so Karlsson initially stands between jars nn and n+1n+1.

What is the minimum number of jars Karlsson is required to empty so that an equal number of full strawberry and blueberry jam jars is left?

Your program should answer tt independent test cases.

Input

The first line contains one integer tt (1≤t≤10001≤t≤1000) — the number of test cases.

The first line of each test case contains a single integer nn (1≤n≤1051≤n≤105).

The second line of each test case contains 2n2n integers a1,a2,…,a2na1,a2,…,a2n (1≤ai≤21≤ai≤2) — ai=1ai=1 means that the ii-th jar from the left is a strawberry jam jar and ai=2ai=2 means that it is a blueberry jam jar.

It is guaranteed that the sum of nn over all test cases does not exceed 105105.

Output

For each test case print the answer to it — the minimum number of jars Karlsson is required to empty so that an equal number of full strawberry and blueberry jam jars is left.

Example

input

Copy

4
6
1 1 1 2 2 1 2 1 2 1 1 2
2
1 2 1 2
3
1 1 1 1 1 1
2
2 1 1 1

output

Copy

6
0
6
2

Note

The picture from the statement describes the first test case.

In the second test case the number of strawberry and blueberry jam jars is already equal.

In the third test case Karlsson is required to eat all 66 jars so that there remain 00 jars of both jams.

In the fourth test case Karlsson can empty either the second and the third jars or the third and the fourth one. The both scenarios will leave 11 jar of both jams.

 

 

#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
//#include <unordered_map>
#include <map>
#include <queue>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
//#include <multimap>
#define ll long long
#define inf 0x3f3f3f3
using namespace std;
const int mxn = 2e5+10;
#define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ls now<<1,l,mid
#define rs now<<1|1,mid+1,r
#define lc now<<1
#define rc now<<1|1
#define upsum(now) rt[now].sum =rt[now<<1].sum + rt[now<<1|1].sum ;
#define upmx(now) rt[now].mx = max(rt[now<<1].mx , rt[now<<1|1].mx) ;
#define pb push_back
int n,m,k,t,mx,mn,l,r,dp[mxn],u,v,cost;
int l1[mxn>>1],l2[mxn>>1], r1[mxn>>1],r2[mxn>>1];
string str,ch;
int main()
{
    TLE;
    cin>>t;
    for(int tt=1; tt<=t; tt++)
    {
        cin>>n;
        map<int,int>mp;
        int ans = 0 ;
        mp[0] = n ;
        for(int i=0;i<n;i++)
        {
            cin>>dp[i];
            ans += ( dp[i]==1? -1: 1 );
            mp[ans] = n-1-i;
        }
        for(int i=n-1;i>=0;i--)
            cin>>dp[i];
        ans = 0 ; mn =mp[0]+n;
        for(int i=0;i<n;i++)
        {
            ans += ( dp[i]==1? -1: 1 );
            if(mp.count(-ans))
            {
                mn = min( mn , mp[-ans]+n-1-i );
            }
        }
        cout<<mn<<endl;
    }
    return 0 ;
}