def solve():
n = int(input())
a = list(map(int, input().split()))
b = [0] * 10
for i in range(n):
b[a[i] % 10] += 1
ans = 'no'
for i in range(10):
for j in range(10):
for k in range(10):
if (i + j + k) % 10 != 3:
continue
b[i] -= 1
b[j] -= 1
b[k] -= 1
if b[i] >= 0 and b[j] >= 0 and b[k] >= 0:
ans = 'yes'
b[i] += 1
b[j] += 1
b[k] += 1
print(ans)
def main():
t = int(input())
for _ in range(t):
solve()
main()
#include <bits/stdc++.h>
using namespace std;
void solve(vector<int> v)
{
for (int i = 0; i < v.size() - 2; i++)
{
for (int j = i + 1; j < v.size() - 1; j++)
{
for (int k = j + 1; k < v.size(); k++)
{
if ((v[i] + v[j] + v[k]) % 10 == 3)
{
cout << "YES" << endl;
return;
}
}
}
}
cout << "NO" << endl;
}
int main()
{
int t;
cin >> t;
while (t--)
{
int n;
cin >> n;
map<int, int> mp;
for (int i = 0; i < n; i++)
{
long long int data;
cin >> data;
int r = data % 10;
mp[r]++;
}
map<int, int>::iterator it = mp.begin();
// Iterate through the map and print the elements
vector<int> v;
while (it != mp.end())
{
if (it->second >= 3)
{
for (int i = 0; i < 3; i++)
{
v.push_back(it->first);
}
}
else if (it->second <= 2)
{
for (int i = 0; i < it->second; i++)
{
v.push_back(it->first);
}
}
++it;
}
solve(v);
}
return 0;
}580A- Kefa and First Steps | 1472B- Fair Division |
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590. N-ary Tree Postorder Traversal | 589. N-ary Tree Preorder Traversal |
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561. Array Partition I | 1374. Generate a String With Characters That Have Odd Counts |