/*
JAI JAGANNATH!
*/
//@Author : zanj0
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = __gnu_pbds::tree<T, __gnu_pbds::null_type, less<T>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update>;
// #define LOCAL // ← enable locally via -DLOCAL; keep commented for OJ
#define ff first
#define ss second
#define pb push_back
#define MOD 1000000007
#define inf 1000000000000000000LL
#define ps(x, y) fixed << setprecision(y) << x
#define w(x) \
int x; \
cin >> x; \
while (x--)
#define endl "\n"
#define timetaken cerr << "Time : " << 1000 * (long double)clock() / (long double)CLOCKS_PER_SEC << "ms\n"
typedef long long int lli;
#ifdef LOCAL
#define dbg(x) cerr << "[DBG] " << #x << " = " << (x) << '\n'
template <class A, class B>
ostream &operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ',' << p.second << ')'; }
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v)
{
os << '[';
for (size_t i = 0; i < v.size(); ++i)
{
if (i)
os << ',';
os << v[i];
}
return os << ']';
}
template <class K, class V>
ostream &operator<<(ostream &os, const map<K, V> &mp)
{
os << '{';
bool first = true;
for (const auto &kv : mp)
{
if (!first)
os << ',';
first = false;
os << kv.first << ':' << kv.second;
}
return os << '}';
}
template <class K, class V>
ostream &operator<<(ostream &os, const unordered_map<K, V> &mp)
{
os << '{';
bool first = true;
for (const auto &kv : mp)
{
if (!first)
os << ',';
first = false;
os << kv.first << ':' << kv.second;
}
return os << '}';
}
#else
#define dbg(x) ((void)0)
#endif
void zanj0()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
#ifdef LOCAL
freopen("input.txt", "r", stdin);
#endif
}
/*
────────────────────────────────────────────────────────────────────────
Problem Statement:
Observations:
1 1 2 2
Sum of element picked <= n -> maximise the sum.
Knapsack?
Claims:
────────────────────────────────────────────────────────────────────────
*/
void Solve()
{
lli n;
cin >> n;
vector<lli> v(n);
for (int i = 0; i < n; i++)
{
cin >> v[i];
}
vector<bool> dp(n + 1);
dp[0] = true;
for (auto &it : v)
{
for (lli j = n - it; j >= 0; j--)
{
if (dp[j])
dp[j + it] = true;
}
}
lli ret = 0;
for (int i = n; i >= 0; i--)
{
if (dp[i])
{
ret = i;
break;
}
}
cout << ret << endl;
}
int32_t main()
{
zanj0();
w(t) Solve();
timetaken;
return 0;
}
/*
GOLDEN RULES
• Solutions are simple.
• Proofs are simple.
• Implementations are simple.
*/