Problem: Problem

There are numbers from 0 to N-1. I would like to select the lucky number for M + 1 days by the following methods. Randomly decide the lucky number on the first day. If the lucky number after i days is A and the lucky number after (i + 1) days is B,


B = (A + j)% N
(However, j is all integers where 0 ≤ j <N and (j / K) is an even number.)


Randomly decide from one of B required in. However, the result of a / b is an integer rounded down to the nearest whole number, and a% b is the remainder of a divided by b. K is also guaranteed to be a number that is divisible by N / K and N / K is even.

For example, when there are numbers 0,1,2,3 and K = 1.

* The lucky number next to 0 is 0 or 2
* The lucky number next to 1 is 1 or 3
* The lucky number next to 2 is 0 or 2
* The next lucky number after 3 is 1 or 3



It will be.

Then Q questions are given. The content of each question is as follows.

When the lucky number on the first day is a, how can you choose the lucky number up to b days so that the lucky number after b days becomes c? There will be a huge number of choices, so find the remainder after dividing by 1,000,000,007.

For example, if N = 4 K = 1, the first lucky number is 0 and the lucky number after 3 days is 2.

* 0-> 0-> 0-> 2
* 0-> 0-> 2-> 2
* 0-> 2-> 0-> 2
* 0-> 2-> 2-> 2



There are four ways.

Constraints

* 1 ≤ N ≤ 5000
* 1 ≤ M ≤ 5000
* 1 ≤ K ≤ 5000
* N / K is divisible
* N / K is even
* 1 ≤ Q ≤ 100000
* 0 ≤ ai <N (1 ≤ i ≤ Q)
* 0 ≤ bi ≤ M (1 ≤ i ≤ Q)
* 0 ≤ ci <N (1 ≤ i ≤ Q)

Input

The input is given in the following format.


N M K Q
a1 b1 c1
a2 b2 c2
..
..
..
aQ bQ cQ


Output

For each question, ask for an answer and output the remainder divided by 1,000,000,007 line by line.

Example

Input

6 3 1 10
0 1 0
0 2 0
0 3 0
1 3 3
0 2 1
2 2 2
2 3 2
2 2 0
1 1 1
2 2 1


Output

1
3
9
9
0
3
9
3
1
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n,m,K,q;
ll dp[5005][5005];
ll tp[5005];

ll get(int f,int t){
	if(f>=0){
		return (tp[t]-tp[f]+MOD)%MOD;
	}else{
		return (tp[K*2]-tp[K*2+f]+tp[t]+MOD)%MOD;
	}
}

int main(void){
	scanf("%d%d%d%d",&n,&m,&K,&q);
	dp[0][0]=1;
	for(int i=0;i<=m;i++){
		memset(tp,0,sizeof(tp));
		for(int j=0;j<2*K;j++){
			for(int k=0;k<n/(2*K);k++){
				tp[j+1]+=dp[i][k*(2*K)+j];
				tp[j+1]%=MOD;
			}
		}
		for(int j=0;j<2*K;j++){
			tp[j+1]+=tp[j];
			tp[j+1]%=MOD;
		}
		for(int j=0;j<2*K;j++){
			for(int k=0;k<n/(2*K);k++){
				//printf("%d %d %d %lld\n",i,j,k,get(j,j+K));
				dp[i+1][k*(2*K)+j]=get(j-K+1,j+1);
				//printf("%d %d %d %d %lld\n",i+1,k*(2*K)+j,j-K+1,j+1,dp[i+1][k*(2*K)+j]);
			}
		}
	}
	for(int i=0;i<q;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		int dif=(c-a+n)%n;
		printf("%lld\n",dp[b][dif]);
	}
	return 0;
}