asyncio 协程并发发送网络请求，如何优先继续处理返回结果的任务，而不是开始发送新的请求？

import asyncio

import httpx

request_time = 10

limits = httpx.Limits(max_connections=5)
client = httpx.AsyncClient(limits=limits)


async def request_baidu(i):
    # async with httpx.AsyncClient(limits=limits) as client:
        print(f"===> %d" % i)
        r = await client.get("http://www.baidu.com")
        # print(r.status_code)
        print(f"<=== %d" % i) # 如何保证优先执行此处代码，而不发送新的请求？


async def main():
    request_list = [asyncio.create_task(request_baidu(i)) for i in range(request_time)]
    await asyncio.gather(*request_list)

if __name__ == '__main__':
    asyncio.run(main())

"""
===> 0
===> 1
===> 2
===> 3
===> 4
===> 5
===> 6
===> 7
===> 8
===> 9
<=== 3
<=== 4
<=== 0 # 可以看到发送完所有的网络请求，才开始处理返回的结果
<=== 1
<=== 2
<=== 5
<=== 6
<=== 7
<=== 8
<=== 9

期望的结果：
===> 0
===> 1
<=== 0 # 在前面的请求放回结果的时候优先返回该处继续向下执行，而不是发送新的请求
===> 2
===> 3
<=== 1
<=== 2
===> 4
...
"""