AtCoder Grand Contest 031
F
Walk on Graph
AtCoder
⌀ 3887
原题
← E→
翻译中…

Problem Statement

翻译中…

You are given a connected graph with $N$ vertices and $M$ edges. The vertices are numbered $1$ to $N$. The $i$-th edge is an undirected edge of length $C_i$ connecting Vertex $A_i$ and Vertex $B_i$.

翻译中…

Additionally, an odd number $MOD$ is given.

翻译中…

You will be given $Q$ queries, which should be processed. The queries take the following form:

翻译中…
  • Given in the $i$-th query are $S_i$, $T_i$ and $R_i$. Print YES if there exists a path from Vertex $S_i$ to Vertex $T_i$ whose length is $R_i$ modulo $MOD$, and print NO otherwise. A path may traverse the same edge multiple times, or go back using the edge it just used.
翻译中…

Here, in this problem, the length of a path is NOT the sum of the lengths of its edges themselves, but the length of the first edge used in the path gets multiplied by $1$, the second edge gets multiplied by $2$, the third edge gets multiplied by $4$, and so on. (More formally, let $L_1,...,L_k$ be the lengths of the edges used, in this order. The length of that path is the sum of $L_i \times 2^{i-1}$.)

翻译中…

Constraints

翻译中…
  • $1 \leq N,M,Q \leq 50000$
  • $3 \leq MOD \leq 10^{6}$
  • $MOD$ is odd.
  • $1 \leq A_i,B_i\leq N$
  • $0 \leq C_i \leq MOD-1$
  • $1 \leq S_i,T_i \leq N$
  • $0 \leq R_i \leq MOD-1$
  • The given graph is connected. (It may contain self-loops or multiple edges.)
翻译中…

Input

翻译中…

Input is given from Standard Input in the following format:

翻译中…

N M Q MOD
A_1 B_1 C_1
\vdots
A_M B_M C_M
S_1 T_1 R_1
\vdots
S_Q T_Q R_Q

翻译中…

Output

翻译中…

Print the answers to the $i$-th query in the $i$-th line.

翻译中…

样例

Input #1
3 2 2 2019
1 2 1
2 3 2
1 3 5
1 3 4
Output #1
YES
NO
The answer to each query is as follows: - The first query: If we take the path $1,2,3$, its length is $1 \times 2^0 + 2 \times 2^1 = 5$, so there exists a path whose length is $5$ modulo $2019$. The answer is `YES`. - The second query: No matter what path we take from Vertex $1$ to Vertex $3$, its length will never be $4$ modulo $2019$. The answer is `NO`.
Input #2
6 6 3 2019
1 2 4
2 3 4
3 4 4
4 5 4
5 6 4
6 1 4
2 6 1110
3 1 1111
4 5 1112
Output #2
YES
NO
NO
Input #3
1 2 3 25
1 1 1
1 1 2
1 1 13
1 1 6
1 1 14
Output #3
YES
YES
YES
Input #4
10 15 10 15
1 2 1
2 3 6
3 4 6
2 5 1
5 6 1
4 7 6
1 8 11
2 9 6
5 10 11
9 10 11
3 6 1
2 5 1
2 7 11
9 10 11
5 6 11
1 3 5
9 8 3
7 7 7
7 10 13
4 1 10
9 3 12
10 10 14
9 2 1
6 6 5
8 8 4
Output #4
YES
NO
NO
NO
NO
NO
NO
YES
YES
NO
题目信息
题号
agc031_f
时间限制
2 s
内存限制
1024 MB

题面与样例版权归 AtCoder 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。