Problem: And now the numerous qualifying tournaments for one of the most prestigious Russian contests Russian Codec Cup are over. All n participants who have made it to the finals found themselves in a huge m-floored 108-star hotel. Of course the first thought to come in a place like this is "How about checking out the elevator?".

The hotel's elevator moves between floors according to one never changing scheme. Initially (at the moment of time 0) the elevator is located on the 1-st floor, then it moves to the 2-nd floor, then — to the 3-rd floor and so on until it reaches the m-th floor. After that the elevator moves to floor m - 1, then to floor m - 2, and so on until it reaches the first floor. This process is repeated infinitely. We know that the elevator has infinite capacity; we also know that on every floor people get on the elevator immediately. Moving between the floors takes a unit of time.

For each of the n participant you are given si, which represents the floor where the i-th participant starts, fi, which represents the floor the i-th participant wants to reach, and ti, which represents the time when the i-th participant starts on the floor si.

For each participant print the minimum time of his/her arrival to the floor fi. 

If the elevator stops on the floor si at the time ti, then the i-th participant can enter the elevator immediately. If the participant starts on the floor si and that's the floor he wanted to reach initially (si = fi), then the time of arrival to the floor fi for this participant is considered equal to ti.

Input

The first line contains two space-separated integers n and m (1 ≤ n ≤ 105, 2 ≤ m ≤ 108). 

Next n lines contain information about the participants in the form of three space-separated integers si fi ti (1 ≤ si, fi ≤ m, 0 ≤ ti ≤ 108), described in the problem statement.

Output

Print n lines each containing one integer — the time of the arrival for each participant to the required floor.

Examples

Input

7 4
2 4 3
1 2 0
2 2 0
1 2 1
4 3 5
1 2 2
4 2 0


Output

9
1
0
7
10
7
5


Input

5 5
1 5 4
1 3 1
1 3 4
3 1 5
4 2 5


Output

12
10
10
8
7

Note

Let's consider the first sample. The first participant starts at floor s = 2 by the time equal to t = 3. To get to the floor f = 4, he has to wait until the time equals 7, that's the time when the elevator will go upwards for the second time. Then the first participant should get on the elevator and go two floors up. In this case the first participant gets to the floor f at time equal to 9. The second participant starts at the time t = 0 on the floor s = 1, enters the elevator immediately, and arrives to the floor f = 2. The third participant doesn't wait for the elevator, because he needs to arrive to the same floor where he starts.

Solution (in JAVA):

Solution:
import java.awt.Point;
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import static java.lang.Math.*;
 
public class Solution117A {
        
        final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE")!=null;
        BufferedReader in;
        PrintWriter out;
        StringTokenizer tok = new StringTokenizer("");
        
        void init() throws FileNotFoundException{
                if (ONLINE_JUDGE){
                        in = new BufferedReader(new InputStreamReader(System.in));
                        out = new PrintWriter(System.out);
                }else{
                        in = new BufferedReader(new FileReader("input.txt"));
                        out = new PrintWriter("output.txt");
                }
        }
        
        String readString() throws IOException{
                while(!tok.hasMoreTokens()){
                        tok = new StringTokenizer(in.readLine());
                }
                return tok.nextToken();
        }
        
        int readInt() throws IOException{
                return Integer.parseInt(readString());
        }
        
        long readLong() throws IOException{
                return Long.parseLong(readString());
        }
        
        double readDouble() throws IOException{
                return Double.parseDouble(readString());
        }
        
        public static void main(String[] args){
                new Solution117A().run();
        }
        
        public void run(){
                try{
                        long t1 = System.currentTimeMillis();
                        init();
                        solve();
                        out.close();
                        long t2 = System.currentTimeMillis();
                        System.err.println("Time = "+(t2-t1));
                }catch (Exception e){
                        e.printStackTrace(System.err);
                        System.exit(-1);
                }
        }
        
        void solve() throws IOException{
        	int n = readInt();
        	int m = readInt()-1;
        	int[] s = new int[n];
        	int[] f = new int[n];
        	int[] t = new int[n];
        	for(int i = 0; i < n; i++){
        		s[i] = readInt()-1;
        		f[i] = readInt()-1;
        		t[i] = readInt();
        	}
        	int mod = m*2;
        	for(int i = 0; i < n; i++){
        		if(s[i] == f[i]){
        			out.println(t[i]);
        			continue;
        		}
        		int curT = t[i];
        		if(s[i] > f[i]){
        			s[i] = m + (m - s[i]);
            		f[i] = m + (m - f[i]);
        		}
    			if(s[i] < t[i] % mod){
    				curT += mod - t[i] % mod + s[i];
    				curT += f[i]-s[i];
    				out.println(curT);
    			}
    			else{
    				curT += (s[i] - t[i] % mod);
    				curT += f[i]-s[i];
    				out.println(curT);
    			}
        	}
        }
        
        static class Utils {
 
                private Utils() {}
 
                public static void mergeSort(int[] a) {
                        mergeSort(a, 0, a.length - 1);
                }
 
                private static void mergeSort(int[] a, int leftIndex, int rightIndex) {
                        final int MAGIC_VALUE = 50;
                        if (leftIndex < rightIndex) {
                                if (rightIndex - leftIndex <= MAGIC_VALUE) {
                                        insertionSort(a, leftIndex, rightIndex);
                                } else {
                                        int middleIndex = (leftIndex + rightIndex) / 2;
                                        mergeSort(a, leftIndex, middleIndex);
                                        mergeSort(a, middleIndex + 1, rightIndex);
                                        merge(a, leftIndex, middleIndex, rightIndex);
                                }
                        }
                }
 
                private static void merge(int[] a, int leftIndex, int middleIndex, int rightIndex) {
                        int length1 = middleIndex - leftIndex + 1;
                        int length2 = rightIndex - middleIndex;
                        int[] leftArray = new int[length1];
                        int[] rightArray = new int[length2];
                        System.arraycopy(a, leftIndex, leftArray, 0, length1);
                        System.arraycopy(a, middleIndex + 1, rightArray, 0, length2);
                        for (int k = leftIndex, i = 0, j = 0; k <= rightIndex; k++) {
                                if (i == length1) {
                                        a[k] = rightArray[j++];
                                } else if (j == length2) {
                                        a[k] = leftArray[i++];
                                } else {
                                        a[k] = leftArray[i] <= rightArray[j] ? leftArray[i++] : rightArray[j++];
                                }
                        }
                }
 
                private static void insertionSort(int[] a, int leftIndex, int rightIndex) {
                        for (int i = leftIndex + 1; i <= rightIndex; i++) {
                                int current = a[i];
                                int j = i - 1;
                                while (j >= leftIndex && a[j] > current) {
                                        a[j + 1] = a[j];
                                        j--;
                                }
                                a[j + 1] = current;
                        }
                } 
        }
        
       
        
        boolean isPrime(int a){
        	for(int i = 2; i <= sqrt(a); i++)
        		if(a % i == 0) return false;
        	return true;
        }
        
        static double distance(long x1, long y1, long x2, long y2){
        	return Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
        }
        
        static long gcd(long a, long b){
        	if(min(a,b) == 0) return max(a,b);
        	return gcd(max(a, b) % min(a,b), min(a,b));
        }
        
        static long lcm(long a, long b){
        	return a * b /gcd(a, b);
        }
}