Problem: Misha walked through the snowy forest and he was so fascinated by the trees to decide to draw his own tree!

Misha would like to construct a rooted tree with n vertices, indexed from 1 to n, where the root has index 1. Every other vertex has a parent p_i, and i is called a child of vertex p_i. Vertex u belongs to the subtree of vertex v iff v is reachable from u while iterating over the parents (u, p_{u}, p_{p_{u}}, ...). Clearly, v belongs to its own subtree, and the number of vertices in the subtree is called the size of the subtree. Misha is only interested in trees where every vertex belongs to the subtree of vertex 1.

Below there is a tree with 6 vertices. The subtree of vertex 2 contains vertices 2, 3, 4, 5. Hence the size of its subtree is 4. 

<image>

The branching coefficient of the tree is defined as the maximum number of children in any vertex. For example, for the tree above the branching coefficient equals 2. Your task is to construct a tree with n vertices such that the sum of the subtree sizes for all vertices equals s, and the branching coefficient is minimum possible.

Input

The only input line contains two integers n and s — the number of vertices in the tree and the desired sum of the subtree sizes (2 ≤ n ≤ 10^5; 1 ≤ s ≤ 10^{10}).

Output

If the required tree does not exist, output «No». Otherwise output «Yes» on the first line, and in the next one output integers p_2, p_3, ..., p_n, where p_i denotes the parent of vertex i.

Examples

Input


3 5


Output


Yes
1 1 


Input


4 42


Output


No


Input


6 15


Output


Yes
1 2 3 1 5 

Note

Below one can find one of the possible solutions for the first sample case. The sum of subtree sizes equals 3 + 1 + 1 = 5, and the branching coefficient equals 2.

<image>

Below one can find one of the possible solutions for the third sample case. The sum of subtree sizes equals 6 + 3 + 2 + 1 + 2 + 1 = 15, and the branching coefficient equals 2.

<image>

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
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;
		FastScanner in = new FastScanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskC solver = new TaskC();
		solver.solve(1, in, out);
		out.close();
	}

	static class TaskC {
		public void solve(int testNumber, FastScanner in, PrintWriter out) {
			int n = in.nextInt();
			long s = in.nextLong();
			if (s < 2 * n - 1 || s > (long) n * (n + 1) / 2) {
				out.println("No");
				return;
			}

			int l = 0;
			int r = n - 1;
			while (r - l > 1) {
				int m = (l + r) / 2;
				int[] p = new int[n];
				p[0] = -1;
				int[] nc = new int[n];
				int last = 0;
				for (int i = 1; i < n; i++) {
					while (nc[last] >= m) {
						++last;
					}
					if (last >= i) {
						throw new AssertionError();
					}
					++nc[last];
					p[i] = last;
				}
				if (sumDepths(p) <= s) {
					r = m;
				} else {
					l = m;
				}
			}

			int[] p = new int[n];
			p[0] = -1;
			int maxDeg = r;

			long need = (long) n * (n + 1) / 2 - s;
			int[] nc = new int[n];
			int[] d = new int[n];
			d[0] = 1;
			for (int i = 1; i < n; i++) {
				p[i] = i - 1;
				++nc[p[i]];
				d[i] = 1 + d[p[i]];
			}

			List<Integer>[] byDepth = new List[n + 1];
			for (int i = 0; i <= n; i++) {
				byDepth[i] = new ArrayList<>();
			}
			for (int i = 0; i < n; i++) {
				byDepth[d[i]].add(i);
			}
			int j = 1;
			for (int i = n - 1; i >= 0; i--) {
				if (need == 0) {
					break;
				}

				int newParent = -1;
				while (true) {
					while (byDepth[j].isEmpty()) {
						++j;
					}
					int x = byDepth[j].get(byDepth[j].size() - 1);
					if (d[x] != j || nc[x] == maxDeg) {
						byDepth[j].remove(byDepth[j].size() - 1);
						continue;
					}
					int delta = d[i] - (j + 1);
					if (delta > need) {
						++j;
						continue;
					}
					newParent = x;
					break;
				}

				--nc[p[i]];
				need -= d[i];
				p[i] = newParent;
				d[i] = 1 + d[newParent];
				++nc[p[i]];
				need += d[i];
				byDepth[d[i]].add(i);
			}

//		if (sumDepths(p) != s) {
//			throw new AssertionError(sumDepths(p) + " " + s + " " + Arrays.toString(p));
//		}
			out.println("Yes");
			for (int i = 1; i < n; i++) {
				if (i > 1) {
					out.print(" ");
				}
				out.print(p[i] + 1);
			}
			out.println();
		}

		private long sumDepths(int[] p) {
			int n = p.length;
			int[] d = new int[n];
			d[0] = 1;
			long res = 1;
			for (int i = 1; i < n; i++) {
				d[i] = 1 + d[p[i]];
				res += d[i];
			}
			return res;
		}

	}

	static class FastScanner {
		private BufferedReader in;
		private StringTokenizer st;

		public FastScanner(InputStream stream) {
			in = new BufferedReader(new InputStreamReader(stream));
		}

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

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

		public long nextLong() {
			return Long.parseLong(next());
		}

	}
}