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>
using namespace std;
using Int = long long;
const int MOD = 1e9+7;
int sm[10050];
int dp[5050][5050];
signed main(){  
  cin.tie(0);
  ios::sync_with_stdio(0);
  int n,m,k,q;
  cin>>n>>m>>k>>q;
  memset(dp,0,sizeof(dp));
  dp[0][0]=1;
  for(int i=0;i<m;i++){
    memset(sm,0,sizeof(sm));
    for(int j=0;j<n;j++){
      int x=j%(2*k);
      sm[x]+=dp[i][j];
      sm[x]%=MOD;
      sm[2*k+x]+=dp[i][j];
      sm[2*k+x]%=MOD;
    }
    
    for(int j=1;j<4*k;j++) sm[j]=(sm[j]+sm[j-1])%MOD;
      
    for(int j=0;j<n;j++)
      dp[i+1][j]=(sm[2*k+(j%(2*k))]+MOD-sm[k+(j%(2*k))])%MOD;
  }
  
  for(int i=0;i<q;i++){
    int a,b,c;
    cin>>a>>b>>c;
    int d=(c-a+n)%n;
    cout<<dp[b][d]<<endl;
  }
  
  return 0;
}