Let \(S\) be the set of deer that ride on the sled. Then the condition is

\[\displaystyle \sum_{i\notin S}P_i\geq \sum_{i \in S}W_i.\]

Plugging \(\displaystyle \sum_{i\notin S}P_i=\sum_{i=1}^N P_i - \sum_{i \in S}P_i\), we obtain

\[\displaystyle \sum_{i\in S}(W_i+P_i)\leq \sum_{i=1}^N P_i.\]

Since the right hand side is a constant, it is optimal to put deer in ascending order of \(W_i+P_i\) greedily.

The computational complexity is \(O(N\log N)\), where the bottleneck is sorting by \(W_i+P_i\).

def solve():
    n = int(input())
    wp = [list(map(int, input().split())) for i in range(n)]
    wp.sort(key=lambda x: x[0] + x[1]) # Sort by w_i + p_i
    sum_p = 0
    for w, p in wp:
        sum_p += p
    res = 0
    for i in range(n):
        w, p = wp[i]
        res += w + p
        if sum_p < res:
            print(i)
            return


for _ in range(int(input())):
    solve()