Codeforces Round 489 (Div. 2)
E
Nastya and King-Shamans
Codeforces
2500
原题
← D→
翻译中…

Nastya likes reading and even spends whole days in a library sometimes. Today she found a chronicle of Byteland in the library, and it stated that there lived shamans long time ago. It is known that at every moment there was exactly one shaman in Byteland, and there were $*n*$ shamans in total enumerated with integers from $1$ to $*n*$ in the order they lived. Also, each shaman had a magic power which can now be expressed as an integer.

翻译中…

The chronicle includes a list of powers of the $*n*$ shamans. Also, some shamans can be king-shamans, if they gathered all the power of their predecessors, i.e. their power is exactly the sum of powers of all previous shamans. Nastya is interested in whether there was at least one king-shaman in Byteland.

翻译中…

Unfortunately many of the powers are unreadable in the list, so Nastya is doing the following:

翻译中…
  • Initially she supposes some power for each shaman.
  • After that she changes the power of some shaman $*q*$ times (the shamans can differ) and after that wants to check if there is at least one king-shaman in the list. If yes, she wants to know the index of any king-shaman.
翻译中…

Unfortunately the list is too large and Nastya wants you to help her.

翻译中…

Input

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The first line contains two integers $*n*$ and $*q*$ ($1 ≤ *n*, *q* ≤ 2·10^{5}$).

翻译中…

The second line contains $*n*$ integers $*a*_{1}, ..., *a*_{*n*}$ ($0 ≤ *a*_{*i*} ≤ 10^{9}$), where $*a*_{*i*}$ is the magic power of the $*i*$-th shaman.

翻译中…

After that $*q*$ lines follow, the $*i*$-th of them contains two integers $*p*_{*i*}$ and $*x*_{*i*}$ ($1 ≤ *p*_{*i*} ≤ *n*$, $0 ≤ *x*_{*i*} ≤ 10^{9}$) that mean that the new power of the $*p*_{*i*}$-th shaman is $*x*_{*i*}$.

翻译中…

Output

翻译中…

Print $*q*$ lines, the $*i*$-th of them should contain $- 1$, if after the $*i*$-th change there are no shaman-kings, and otherwise a single integer $*j*$, where $*j*$ is an index of some king-shaman after the $*i*$-th change.

翻译中…

If there are multiple king-shamans after each change, print the index of any of them.

翻译中…

Note

翻译中…

In the first example powers of shamans after the first change are equal to $(2, 3)$. The answer equals $- 1$, because the sum of powers of shamans before the first shaman is equal to $0$, and before the second is equal to $2$.

翻译中…

In the second example after the first change the powers are equal to $(1, 2, 3)$. The answer is equal to $3$, because the power of the third shaman is equal to $3$, and the sum of powers of the first and the second shaman is also $1 + 2 = 3$. After the second change the powers become equal to $(2, 2, 3)$, where the answer equals $2$. After the third change the powers become equal to $(2, 4, 3)$, where the answer equals $- 1$. After the fourth change the powers become equal to $(2, 4, 6)$, where the answer equals $3$.

翻译中…

样例

Input #1
2 1
1 3
1 2
Output #1
-1
InputCopy OutputCopy
Input #2
3 4
2 2 3
1 1
1 2
2 4
3 6
Output #2
3
2
-1
3
InputCopy OutputCopy
Input #3
10 7
0 3 1 4 6 2 7 8 10 1
2 5
1 3
9 36
4 10
4 9
1 2
1 0
Output #3
1
-1
9
-1
4
-1
1
InputCopy OutputCopy
题目信息
题号
992E
时间限制
3 s
内存限制
256 MB
标签
二分
数据结构

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。