Problem: There are n cities in the country of Berland. Some of them are connected by bidirectional roads in such a way that there exists exactly one path, which visits each road no more than once, between every pair of cities. Each road has its own length. Cities are numbered from 1 to n.

The travelling time between some cities v and u is the total length of the roads on the shortest path from v to u. 

The two most important cities in Berland are cities 1 and n.

The Berland Ministry of Transport decided to build a single new road to decrease the traffic between the most important cities. However, lots of people are used to the current travelling time between the most important cities, so the new road shouldn't change it too much. 

The new road can only be built between such cities v and u that v ≠ u and v and u aren't already connected by some road.

They came up with m possible projects. Each project is just the length x of the new road.

Polycarp works as a head analyst at the Berland Ministry of Transport and it's his job to deal with all those m projects. For the i-th project he is required to choose some cities v and u to build the new road of length x_i between such that the travelling time between the most important cities is maximal possible. 

Unfortunately, Polycarp is not a programmer and no analyst in the world is capable to process all projects using only pen and paper. 

Thus, he asks you to help him to calculate the maximal possible travelling time between the most important cities for each project. Note that the choice of v and u can differ for different projects.

Input

The first line contains two integers n and m (3 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ m ≤ 3 ⋅ 10^5) — the number of cities and the number of projects, respectively.

Each of the next n - 1 lines contains three integers v_i, u_i and w_i (1 ≤ v_i, u_i ≤ n, 1 ≤ w_i ≤ 10^9) — the description of the i-th road. It is guaranteed that there exists exactly one path, which visits each road no more than once, between every pair of cities.

Each of the next m lines contains a single integer x_j (1 ≤ x_j ≤ 10^9) — the length of the road for the j-th project.

Output

Print m lines, the j-th line should contain a single integer — the maximal possible travelling time between the most important cities for the j-th project.

Example

Input

7 2
1 2 18
2 3 22
3 4 24
4 7 24
2 6 4
3 5 12
1
100


Output

83
88

Note

The road network from the first example:

<image>

You can build the road with length 1 between cities 5 and 6 to get 83 as the travelling time between 1 and 7 (1 → 2 → 6 → 5 → 3 → 4 → 7 = 18 + 4 + 1 + 12 + 24 + 24 = 83). Other possible pairs of cities will give answers less or equal to 83.

Solution (in JAVA):

Solution:
//package educational.round48;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
import java.util.InputMismatchException;

public class F {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		int n = ni(), Q = ni();
		int[] from = new int[n - 1];
		int[] to = new int[n - 1];
		int[] ws = new int[n-1];
		for (int i = 0; i < n - 1; i++) {
			from[i] = ni() - 1;
			to[i] = ni() - 1;
			ws[i] = ni();
		}
		int[][][] g = packWU(n, from, to, ws);
		long[] ds;
		{
			int[][] pars = parents(g, 0);
			int[] par = pars[0], ord = pars[1], dep = pars[2];
			int[] pw = pars[4];
			long[] dw = new long[n];
			for(int i = 1;i < n;i++){
				int cur = ord[i];
				dw[cur] = dw[par[cur]] + pw[cur];
			}
			ds = dw;
		}
		
		int[][] pars = parents(g, n-1);
		int[] par = pars[0], ord = pars[1], dep = pars[2];
		int[] pw = pars[4];
		long[] dw = new long[n];
		for(int i = 1;i < n;i++){
			int cur = ord[i];
			dw[cur] = dw[par[cur]] + pw[cur];
		}
		
		long def = dw[0];
		long[][] us = new long[n][];
		for(int i = 0;i < n;i++){
			us[i] = new long[]{dw[i], i};
		}
		Arrays.sort(us, new Comparator<long[]>() {
			public int compare(long[] a, long[] b) {
				return -Long.compare(a[0], b[0]);
			}
		});
		int[] iord = new int[n];
		for(int i = 0;i < n;i++){
			iord[(int)us[i][1]] = i;
		}
		
		SegmentTreeRMQL st = new SegmentTreeRMQL(n);
		
		long ans = 0;
		for(int i = 0;i < n;i++){
			int cur = (int)us[i][1];
			for(int[] e : g[cur]){
				st.update(e[0], Long.MAX_VALUE / 2);
			}
			if(i > 0)ans = Math.max(ans, us[i][0] + (-st.st[1]));
			for(int[] e : g[cur]){
				if(iord[e[0]] < i){
					st.update(e[0], -ds[e[0]]);
				}
			}
			st.update(cur, -ds[cur]);
		}
		
		for(int z = 0;z < Q;z++){
			int x = ni();
			out.println(Math.min(ans+x, def));
		}
	}
	
	public static class SegmentTreeRMQL {
		public int M, H, N;
		public long[] st;
		
		public SegmentTreeRMQL(int n)
		{
			N = n;
			M = Integer.highestOneBit(Math.max(N-1, 1))<<2;
			H = M>>>1;
			st = new long[M];
			Arrays.fill(st, 0, M, Long.MAX_VALUE);
		}
		
		public SegmentTreeRMQL(long[] a)
		{
			N = a.length;
			M = Integer.highestOneBit(Math.max(N-1, 1))<<2;
			H = M>>>1;
			st = new long[M];
			for(int i = 0;i < N;i++){
				st[H+i] = a[i];
			}
			Arrays.fill(st, H+N, M, Long.MAX_VALUE);
			for(int i = H-1;i >= 1;i--)propagate(i);
		}
		
		public void update(int pos, long x)
		{
			st[H+pos] = x;
			for(int i = (H+pos)>>>1;i >= 1;i >>>= 1)propagate(i);
		}
		
		private void propagate(int i)
		{
			st[i] = Math.min(st[2*i], st[2*i+1]);
		}
		
		public long minx(int l, int r){
			long min = Long.MAX_VALUE;
			if(l >= r)return min;
			while(l != 0){
				int f = l&-l;
				if(l+f > r)break;
				long v = st[(H+l)/f];
				if(v < min)min = v;
				l += f;
			}
			
			while(l < r){
				int f = r&-r;
				long v = st[(H+r)/f-1];
				if(v < min)min = v;
				r -= f;
			}
			return min;
		}
		
		public long min(int l, int r){ return l >= r ? 0 : min(l, r, 0, H, 1);}
		
		private long min(int l, int r, int cl, int cr, int cur)
		{
			if(l <= cl && cr <= r){
				return st[cur];
			}else{
				int mid = cl+cr>>>1;
				long ret = Long.MAX_VALUE;
				if(cl < r && l < mid){
					ret = Math.min(ret, min(l, r, cl, mid, 2*cur));
				}
				if(mid < r && l < cr){
					ret = Math.min(ret, min(l, r, mid, cr, 2*cur+1));
				}
				return ret;
			}
		}
		
		public int firstle(int l, long v) {
			int cur = H+l;
			while(true){
				if(st[cur] <= v){
					if(cur < H){
						cur = 2*cur;
					}else{
						return cur-H;
					}
				}else{
					cur++;
					if((cur&cur-1) == 0)return -1;
					if((cur&1)==0)cur>>>=1;
				}
			}
		}
		
		public int lastle(int l, long v) {
			int cur = H+l;
			while(true){
				if(st[cur] <= v){
					if(cur < H){
						cur = 2*cur+1;
					}else{
						return cur-H;
					}
				}else{
					if((cur&cur-1) == 0)return -1;
					cur--;
					if((cur&1)==1)cur>>>=1;
				}
			}
		}
	}

	
	public static int[][] parents(int[][][] g, int root) {
		int n = g.length;
		int[] par = new int[n];
		Arrays.fill(par, -1);
		int[] dw = new int[n];
		int[] pw = new int[n];
		int[] dep = new int[n];

		int[] q = new int[n];
		q[0] = root;
		for (int p = 0, r = 1; p < r; p++) {
			int cur = q[p];
			for (int[] nex : g[cur]) {
				if (par[cur] != nex[0]) {
					q[r++] = nex[0];
					par[nex[0]] = cur;
					dep[nex[0]] = dep[cur] + 1;
					dw[nex[0]] = dw[cur] + nex[1];
					pw[nex[0]] = nex[1];
				}
			}
		}
		return new int[][] { par, q, dep, dw, pw };
	}

	
	public static int[][][] packWU(int n, int[] from, int[] to, int[] w) {
		int[][][] g = new int[n][][];
		int[] p = new int[n];
		for (int f : from)
			p[f]++;
		for (int t : to)
			p[t]++;
		for (int i = 0; i < n; i++)
			g[i] = new int[p[i]][2];
		for (int i = 0; i < from.length; i++) {
			--p[from[i]];
			g[from[i]][p[from[i]]][0] = to[i];
			g[from[i]][p[from[i]]][1] = w[i];
			--p[to[i]];
			g[to[i]][p[to[i]]][0] = from[i];
			g[to[i]][p[to[i]]][1] = w[i];
		}
		return g;
	}


	public static int[][] parents3(int[][] g, int root) {
		int n = g.length;
		int[] par = new int[n];
		Arrays.fill(par, -1);

		int[] depth = new int[n];
		depth[0] = 0;

		int[] q = new int[n];
		q[0] = root;
		for (int p = 0, r = 1; p < r; p++) {
			int cur = q[p];
			for (int nex : g[cur]) {
				if (par[cur] != nex) {
					q[r++] = nex;
					par[nex] = cur;
					depth[nex] = depth[cur] + 1;
				}
			}
		}
		return new int[][] { par, q, depth };
	}

	static int[][] packU(int n, int[] from, int[] to) {
		int[][] g = new int[n][];
		int[] p = new int[n];
		for (int f : from)
			p[f]++;
		for (int t : to)
			p[t]++;
		for (int i = 0; i < n; i++)
			g[i] = new int[p[i]];
		for (int i = 0; i < from.length; i++) {
			g[from[i]][--p[from[i]]] = to[i];
			g[to[i]][--p[to[i]]] = from[i];
		}
		return g;
	}

	
	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new F().run(); }
	
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}