## Solution:

First of all, we can almost immediately notice that if x[i]>m, then this must have no effect, and we can ignore it directly. Therefore, the value range of x[i] is reduced from 0~$10^9$ to 0~$2\cdot 10^5$.

Further if x[i]<$2\cdot 10^5$ and x[i]+y[i]>m. For this situation, it is very clear, after the addition of i is added, and then after x[i], the answer is permanently increased by 1 (unless it is removed).

Therefore, it is natural to think that, similar to the parenthesis matching problem, mark +1 at now+x[i], mark -1 at now+x[i]+y[i], and use a sum variable to record the sum of +1/-1 marks for current time.

In fact, the above approach has been able to cover the relatively large x[i]+y[i] situation, because if x[i]+y[i] is relatively large, we only need to repeat the mark $\frac m {x[i]+y[i]}$ times.

But if x[i]+y[i] is very small (such as 2), you need to mark $\frac m 2$ times. So it is the time complexity of $m^2$.

For the case where x[i]+y[i] is very small, taking 2 as an example, in fact, it only depends on the time when i is added. Similarly, when x[i]=1 and y[i]=2, we only need to know whether the current time i is under maintenance according to the value of (now-add time)%3.

Therefore, we use $\sqrt m \approx 450$ as the boundary. If it is greater than 450, it is done by pairing the brackets, and if it is less than 450, it can be done by taking the remainder.

## Code:

Java

C++