Let $a$ and $b$ be two arrays of lengths $n$ and $m$, respectively, with no elements in common. We can define a new array $\mathrm{merge}(a,b)$ of length $n+m$ recursively as follows:  If one of the arrays is empty, the result is the other array. That is, $\mathrm{merge}(\emptyset,b)=b$ and $\mathrm{merge}(a,\emptyset)=a$. In particular, $\mathrm{merge}(\emptyset,\emptyset)=\emptyset$.  If both arrays are non-empty, and $a_1<b_1$, then $\mathrm{merge}(a,b)=[a_1]+\mathrm{merge}([a_2,\ldots,a_n],b)$. That is, we delete the first element $a_1$ of $a$, merge the remaining arrays, then add $a_1$ to the beginning of the result.  If both arrays are non-empty, and $a_1>b_1$, then $\mathrm{merge}(a,b)=[b_1]+\mathrm{merge}(a,[b_2,\ldots,b_m])$. That is, we delete the first element $b_1$ of $b$, merge the remaining arrays, then add $b_1$ to the beginning of the result. 
This algorithm has the nice property that if $a$ and $b$ are sorted, then $\mathrm{merge}(a,b)$ will also be sorted. For example, it is used as a subroutine in merge-sort. For this problem, however, we will consider the same procedure acting on non-sorted arrays as well. For example, if $a=[3,1]$ and $b=[2,4]$, then $\mathrm{merge}(a,b)=[2,3,1,4]$.
A permutation is an array consisting of $n$ distinct integers from $1$ to $n$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array) and $[1,3,4]$ is also not a permutation ($n=3$ but there is $4$ in the array).
There is a permutation $p$ of length $2n$. Determine if there exist two arrays $a$ and $b$, each of length $n$ and with no elements in common, so that $p=\mathrm{merge}(a,b)$.
-----Input-----
The first line contains a single integer $t$ ($1\le t\le 1000$)  — the number of test cases. Next $2t$ lines contain descriptions of test cases. 
The first line of each test case contains a single integer $n$ ($1\le n\le 2000$).
The second line of each test case contains $2n$ integers $p_1,\ldots,p_{2n}$ ($1\le p_i\le 2n$). It is guaranteed that $p$ is a permutation.
It is guaranteed that the sum of $n$ across all test cases does not exceed $2000$.
-----Output-----
For each test case, output "YES" if there exist arrays $a$, $b$, each of length $n$ and with no common elements, so that $p=\mathrm{merge}(a,b)$. Otherwise, output "NO".
-----Example-----
Input
6
2
2 3 1 4
2
3 1 2 4
4
3 2 6 1 5 7 8 4
3
1 2 3 4 5 6
4
6 1 3 7 4 5 8 2
6
4 3 2 5 1 11 9 12 8 6 10 7
Output
YES
NO
YES
YES
NO
NO
-----Note-----
In the first test case, $[2,3,1,4]=\mathrm{merge}([3,1],[2,4])$.
In the second test case, we can show that $[3,1,2,4]$ is not the merge of two arrays of length $2$.
In the third test case, $[3,2,6,1,5,7,8,4]=\mathrm{merge}([3,2,8,4],[6,1,5,7])$.
In the fourth test case, $[1,2,3,4,5,6]=\mathrm{merge}([1,3,6],[2,4,5])$, for example.
import sys

input=sys.stdin.readline

for _ in range(int(input())):
    n=int(input())
    p=list(map(int,input().split()))
    index={e:i for i,e in enumerate(p)}
    m=2*n
    data=[]
    for i in range(2*n,0,-1):
        temp=index[i]
        if temp<m:
            data.append(m-temp)
            m=temp
    m=len(data)
    dp=[[False for i in range(2*n+1)] for j in range(m+1)]
    dp[0][0]=True
    for i in range(1,m+1):
        for j in range(2*n+1):
            dp[i][j]=dp[i-1][j]|(dp[i-1][j-data[i-1]]&(j>=data[i-1]))
    if dp[m][n]:
        print("YES")
    else:
        print("NO")

import sys

input=sys.stdin.readline

for _ in range(int(input())):
    n=int(input())
    p=list(map(int,input().split()))
    index={e:i for i,e in enumerate(p)}
    m=2*n
    data=[]
    for i in range(2*n,0,-1):
        temp=index[i]
        if temp<m:
            data.append(m-temp)
            m=temp
    m=len(data)
    dp=[[False for i in range(2*n+1)] for j in range(m+1)]
    dp[0][0]=True
    for i in range(1,m+1):
        for j in range(2*n+1):
            dp[i][j]=dp[i-1][j]|(dp[i-1][j-data[i-1]]&(j>=data[i-1]))
    if dp[m][n]:
        print("YES")
    else:
        print("NO")
