Emuskald was hired to design an artificial waterfall according to the latest trends in landscape architecture. A modern artificial waterfall consists of multiple horizontal panels affixed to a wide flat wall. The water flows down the top of the wall from panel to panel until it reaches the bottom of the wall.

The wall has height t and has n panels on the wall. Each panel is a horizontal segment at height hi which begins at li and ends at ri. The i-th panel connects the points (li, hi) and (ri, hi) of the plane. The top of the wall can be considered a panel connecting the points ( - 109, t) and (109, t). Similarly, the bottom of the wall can be considered a panel connecting the points ( - 109, 0) and (109, 0). No two panels share a common point.

Emuskald knows that for the waterfall to be aesthetically pleasing, it can flow from panel i to panel j (<image>) only if the following conditions hold: 

  1. max(li, lj) < min(ri, rj) (horizontal projections of the panels overlap); 
  2. hj < hi (panel j is below panel i); 
  3. there is no such panel k (hj < hk < hi) that the first two conditions hold for the pairs (i, k) and (k, j). 



Then the flow for <image> is equal to min(ri, rj) - max(li, lj), the length of their horizontal projection overlap.

Emuskald has decided that in his waterfall the water will flow in a single path from top to bottom. If water flows to a panel (except the bottom of the wall), the water will fall further to exactly one lower panel. The total amount of water flow in the waterfall is then defined as the minimum horizontal projection overlap between two consecutive panels in the path of the waterfall. Formally: 

  1. the waterfall consists of a single path of panels <image>; 
  2. the flow of the waterfall is the minimum flow in the path <image>. 



To make a truly great waterfall Emuskald must maximize this water flow, but there are too many panels and he is having a hard time planning his creation. Below is an example of a waterfall Emuskald wants:

<image>

Help Emuskald maintain his reputation and find the value of the maximum possible water flow.

Input

The first line of input contains two space-separated integers n and t (1 ≤ n ≤ 105, 2 ≤ t ≤ 109), the number of the panels excluding the top and the bottom panels, and the height of the wall. Each of the n following lines contain three space-separated integers hi, li and ri (0 < hi < t,  - 109 ≤ li < ri ≤ 109), the height, left and right ends of the i-th panel segment.

It is guaranteed that no two segments share a common point.

Output

Output a single integer — the maximum possible amount of water flow in the desired waterfall.

Examples

Input

5 6
4 1 6
3 2 7
5 9 11
3 10 15
1 13 16


Output

4


Input

6 5
4 2 8
3 1 2
2 2 3
2 6 12
1 0 7
1 8 11


Output

2

Note

The first test case corresponds to the picture.
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.io.PrintStream;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Random;
import java.util.StringTokenizer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskD {
    static class Panel implements Comparable<Panel> {
        int left;
        int right;
        int height;
        int bestFlow = 0;

        public int compareTo(Panel o) {
            return o.height - height;
        }
    }

    static class Tree {
        Tree left;
        Tree right;
        long heapBy;
        int l;
        int r;
        int original;
        int thisOriginal;
        Panel src;

        public Tree leftmost() {
            if (left == null) return this;
            return left.leftmost();
        }

        public Tree rightmost() {
            if (right == null) return this;
            return right.rightmost();
        }

        public int getBestOriginal(boolean skipLeftmost, boolean skipRightmost) {
            if (!skipLeftmost && !skipRightmost)
                return original;
            int res = 0;
            if ((!skipLeftmost || left != null) && (!skipRightmost || right != null))
                res = Math.max(res, thisOriginal);
            if (left != null) res = Math.max(res, left.getBestOriginal(skipLeftmost, false));
            if (right != null) res = Math.max(res, right.getBestOriginal(false, skipRightmost));
            return res;
        }

        public void updateOriginal() {
            original = 0;
            thisOriginal = 0;
            if (src.left == l && src.right == r) {
                thisOriginal = Math.min(r - l, src.bestFlow);
                original = thisOriginal;
            }
            if (left != null) {
                original = Math.max(original, left.original);
            }
            if (right != null) {
                original = Math.max(original, right.original);
            }
        }
    }

    TreePair split(Tree root, int middle) {
        if (root == null) {
            return new TreePair();
        }
        if (root.r <= middle) {
            TreePair pair = split(root.right, middle);
            root.right = pair.left;
            root.updateOriginal();
            pair.left = root;
            return pair;
        } else {
            TreePair pair = split(root.left, middle);
            root.left = pair.right;
            pair.right = root;
            return pair;
        }
    }

    TreePair split2(Tree root, int middle) {
        if (root == null) {
            return new TreePair();
        }
        if (root.l < middle) {
            TreePair pair = split2(root.right, middle);
            root.right = pair.left;
            root.updateOriginal();
            pair.left = root;
            return pair;
        } else {
            TreePair pair = split2(root.left, middle);
            root.left = pair.right;
            root.updateOriginal();
            pair.right = root;
            return pair;
        }
    }

    static class TreePair {
        Tree left;
        Tree right;
    }

    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int t = in.nextInt();
        Panel top = new Panel();
        top.left = (int) -1e9;
        top.right = (int) 1e9;
        top.height = t;
        top.bestFlow = top.right - top.left;
        Panel[] all = new Panel[n + 2];
        all[0] = top;
        Panel bottom = new Panel();
        bottom.left = (int) -1e9;
        bottom.right = (int) 1e9;
        bottom.height = 0;
        all[1] = bottom;
        for (int i = 2; i < all.length; ++i) {
            all[i] = new Panel();
            all[i].height = in.nextInt();
            all[i].left = in.nextInt();
            all[i].right = in.nextInt();
        }
        doit(all);
        out.println(bottom.bestFlow);
    }

    private void doit(Panel[] all) {
        Random r = new Random(System.currentTimeMillis() + 321479371);

        for (int i = 0; i < all.length; ++i) {
            int j = i + r.nextInt(all.length - i);
            Panel tmp = all[j];
            all[j] = all[i];
            all[i] = tmp;
        }
        Arrays.sort(all);
        Tree root = null;
        for (Panel p : all) {
            Tree cur = new Tree();
            cur.l = p.left;
            cur.r = p.right;
            cur.src = p;
            cur.heapBy = r.nextLong();
            cur.updateOriginal();
            if (root == null) {
                root = cur;
                continue;
            }
            TreePair s1 = split(root, cur.l);
            TreePair s2 = split2(s1.right, cur.r);
            Tree untouchedLeft = s1.left;
            Tree untouchedRight = s2.right;
            Tree interesting = s2.left;
            if (interesting.left == null && interesting.right == null) {
                p.bestFlow = Math.max(p.bestFlow, Math.min(p.right - p.left, interesting.src.bestFlow));
                cur.updateOriginal();
                Tree res = untouchedLeft;
                if (interesting.l < cur.l) {
                    Tree nxt = new Tree();
                    nxt.l = interesting.l;
                    nxt.r = cur.l;
                    nxt.src = interesting.src;
                    nxt.heapBy = r.nextLong();
                    nxt.updateOriginal();
                    res = merge(res, nxt);
                }
                res = merge(res, cur);
                if (cur.r < interesting.r) {
                    interesting.l = cur.r;
                    interesting.updateOriginal();
                    res = merge(res, interesting);
                }
                res = merge(res, untouchedRight);
                root = res;
            } else {
                Tree lt = interesting.leftmost();
                if (lt.r == lt.src.right) {
                    p.bestFlow = Math.max(p.bestFlow, Math.min(lt.r - p.left, lt.src.bestFlow));
                }
                Tree rt = interesting.rightmost();
                if (rt.l == rt.src.left) {
                    p.bestFlow = Math.max(p.bestFlow, Math.min(p.right - rt.l, rt.src.bestFlow));
                }
                p.bestFlow = Math.max(p.bestFlow, interesting.getBestOriginal(true, true));
                cur.updateOriginal();
                if (lt.l < cur.l) {
                    lt.left = null;
                    lt.right = null;
                    lt.r = cur.l;
                    lt.updateOriginal();
                    untouchedLeft = merge(untouchedLeft, lt);
                }
                if (cur.r < rt.r) {
                    rt.left = null;
                    rt.right = null;
                    rt.l = cur.r;
                    rt.updateOriginal();
                    untouchedRight = merge(rt, untouchedRight);
                }
                root = merge(untouchedLeft, merge(cur, untouchedRight));
            }
        }
    }

    private Tree merge(Tree left, Tree right) {
        if (left == null) return right;
        if (right == null) return left;
        if (left.heapBy > right.heapBy) {
            left.right = merge(left.right, right);
            left.updateOriginal();
            return left;
        } else {
            right.left = merge(left, right.left);
            right.updateOriginal();
            return right;
        }
    }

    }

class InputReader {
    BufferedReader reader;
    StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    }

