Three from Prostokvashino 2

- Matroskin, I've learned how to build a segment tree. - Sharik, you'd be better off building a new doghouse instead of doing silly things. - Well, Matroskin, I've already shown Uncle Fyodor the code. You can check it in the previous task, and now I'll show you a new function—making updates.

int update(int v, int left, int right, int pos, int val) { if (left == right) t[v] = val; return 0; int mid = (left + right) / 2; if (pos <= mid) update(v * 2, left, mid, pos, val); else update(v * 2 + 1, mid + 1, right, pos, val); t[v] = max(t[v * 2], t[v * 2 + 1]); return 0;}

- Alright, Sharik, since you've written the function, listen. I'll give you an array consisting of $n$ non-negative numbers and a set of queries, each of one of two types: if the query type is one, there will be one number $p$, and you need to find two numbers $l$ and $r$ (1 \leq l \leq r \leq n), such that the function called as follows

get_max(1, 1, n, l, r)

returns a value equal to $p$. If there are several such pairs, output the one with the smallest $l$. If there are still several, output the one with the smallest $r$. If there are no such pairs, output the string "Error". If the query type is two, then two numbers $x$, $y$ follow, and you need to run the function as follows

update(1, 1, n, x, y)

Can you handle such a task? You can assume that at the very beginning, the function

build(1, 1, n)

from the previous task was called.

- I'll try, Matroskin.