import sys
 
def solve():
    # 標準入力から全てのデータを読み込み
    input_tokens = sys.stdin.read().split()
    if not input_tokens:
        return
 
    n = int(input_tokens[0])
 
    # ジョブ情報をリストに格納 [開始時刻, 終了時刻, 価値]
    jobs = []
    idx = 1
    for _ in range(n):
        s = int(input_tokens[idx])
        f = int(input_tokens[idx + 1])
        v = int(input_tokens[idx + 2])
        jobs.append([s, f, v])
        idx += 3
 
    # 1. 終了時刻で安定ソートを実行
    jobs.sort(key=lambda x: x[1])
 
    # 添字を1から扱うため、先頭にダミー要素を追加
    sorted_jobs = [[-1, -1, 0]] + jobs
 
    # 2. q(j) の計算
    q = [0] * (n + 1)
    for j in range(1, n + 1):
        sj = sorted_jobs[j][0]
        for i in range(j - 1, 0, -1):
            if sorted_jobs[i][1] <= sj:
                q[j] = i
                break
 
    # 3. OPT_j の計算 (動的計画法)
    opt = [0] * (n + 1)
    for j in range(1, n + 1):
        vj = sorted_jobs[j][2]
        # 漸化式: OPT_j = max(v_j + OPT_{q(j)}, OPT_{j-1})
        include_j = vj + opt[q[j]]
        exclude_j = opt[j - 1]
 
        if include_j >= exclude_j: # 等しい場合はジョブを含める挙動に合わせる
            opt[j] = include_j
        else:
            opt[j] = exclude_j
 
    # 4. 最適解（ジョブの集合）の計算（バックトラッキング）
    solution = []
    curr = n
    while curr > 0:
        vj = sorted_jobs[curr][2]
        # ジョブcurrを含めた場合の価値が、含めない場合(OPT_{curr-1})以上であれば採用
        if vj + opt[q[curr]] >= opt[curr - 1]:
            solution.append(curr)
            curr = q[curr] # 重ならない直近のジョブへ移動
        else:
            curr -= 1 # ひとつ前のジョブを確認
 
    # 指定された形式（降順リスト）で出力
    print(solution)
 
if __name__ == "__main__":
    solve()

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