2479

/*
DP

30组数据
50000个数，绝对值小于10000
最大连续子段和问题

*/

// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>

#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>

#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <list>
#include <functional>

using namespace std;

// typedef
typedef long long LL;
typedef unsigned long long ULL;

// 
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#define FORi(a,b,c) for(int i=(a);i<(b);i+=c)
#define FORj(a,b,c) for(int j=(a);j<(b);j+=c)
#define FORk(a,b,c) for(int k=(a);k<(b);k+=c)
#define FORp(a,b,c) for(int p=(a);p<(b);p+=c)
#define FORii(a,b,c) for(int ii=(a);ii<(b);ii+=c)
#define FORjj(a,b,c) for(int jj=(a);jj<(b);jj+=c)
#define FORkk(a,b,c) for(int kk=(a);kk<(b);kk+=c)

#define FF(i,a)    for(int i=0;i<(a);i++)
#define FFD(i,a)   for(int i=(a)-1;i>=0;i--)

#define Z(a) (a<<1)
#define Y(a) (a>>1)

const double eps = 1e-6;
const double INFf = 1e100;
const int INFi = 1000000000;
const LL INFll = (LL)1<<62;
const double Pi = acos(-1.0);

template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T TMAX(T x,T y)
{
	if(x>y) return x;
	return y;
}
template<class T> inline T TMIN(T x,T y)
{
	if(x<y) return x;
	return y;
}
template<class T> inline void SWAP(T &x,T &y)
{
	T t = x;
	x = y;
	y = t;
}
template<class T> inline T MMAX(T x,T y,T z)
{
	return TMAX(TMAX(x,y),z);
}
template<class T> inline T MMIN(T x,T y,T z)
{
	return TMIN(TMIN(x,y),z);
}


// code begin
int T,N;
int data[50110];
int DP1[50110];
int DP2[50110];
int main()
{
	read;
	write;
	int sum;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&N);
		FORi(1,N+1,1)
			scanf("%d",data+i);
		DP1[1] = data[1];
		sum = DP1[1];
		FORi(2,N+1,1)
		{
			sum += data[i];
			DP1[i] = TMAX(DP1[i-1],sum);
			if(sum<0) sum =0;
		}
		
		DP2[N] = data[N];
		sum = data[N];
		int ans = DP1[N-1]+DP2[N];
		for(int i=N-1;i>1;i--)
		{
			sum += data[i];
			DP2[i] = TMAX(DP2[i+1],sum);
			if(sum<0) sum = 0;
			if( DP2[i]+DP1[i-1]>ans)
				ans=DP2[i]+DP1[i-1];
		}

		printf("%d\n",ans);
	}
	return 0;
}