Problem: When you were a child you must have been told a puzzle of bags and coins. Anyway, here's one of its versions: 

A horse has three bags. The first bag has one coin, the second bag has one coin and the third bag has three coins. In total, the horse has three coins in the bags. How is that possible?

The answer is quite simple. The third bag contains a coin and two other bags. 

This problem is a generalization of the childhood puzzle. You have n bags. You know that the first bag contains a1 coins, the second bag contains a2 coins, ..., the n-th bag contains an coins. In total, there are s coins. Find the way to arrange the bags and coins so that they match the described scenario or else state that it is impossible to do.

Input

The first line contains two integers n and s (1 ≤ n, s ≤ 70000) — the number of bags and the total number of coins. The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 70000), where ai shows the number of coins in the i-th bag.

Output

If the answer doesn't exist, print -1. 

Otherwise, print n lines, on the i-th line print the contents of the i-th bag. The first number in the line, ci (0 ≤ ci ≤ ai), must represent the number of coins lying directly in the i-th bag (the coins in the bags that are in the i-th bag are not taken into consideration). The second number in the line, ki (0 ≤ ki < n) must represent the number of bags that lie directly in the i-th bag (the bags that are inside the bags lying in the i-th bag are not taken into consideration). Next, the line must contain ki integers — the numbers of the bags that are lying directly in the i-th bag.

The total number of coins in the solution must equal s. If we count the total number of coins the i-th bag in the solution has, we should get ai. 

No bag can directly lie in more than one bag. The bags can be nested in more than one level (see the second test case). If there are multiple correct answers, you can print any of them.

Examples

Input

3 3
1 3 1


Output

1 0
1 2 3 1
1 0


Input

3 3
1 3 1


Output

1 0
2 1 3
0 1 1


Input

1 2
1


Output

-1


Input

8 10
2 7 3 4 1 3 1 2


Output

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

Note

The pictures below show two possible ways to solve one test case from the statement. The left picture corresponds to the first test case, the right picture corresponds to the second one.

<image>

Solution (in JAVA):

Solution:
//package round207;
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 D {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		int n = ni(), s = ni();
		int[] a = na(n);
		int max = -1;
		int maxi = -1;
		for(int i = 0;i < n;i++){
			if(a[i] > max){
				max = a[i];
				maxi = i;
			}
		}
		
		int ss = s - max;
		if(ss < 0){
			out.println(-1);
			return;
		}
		
		int q = (ss>>>6)+1;
		long[] dp = new long[q];
		dp[0] = 1L;
		int[] last = new int[ss+64];
		Arrays.fill(last, -1);
		for(int i = 0;i < n;i++){
			if(maxi == i)continue;
			long[] pre = Arrays.copyOf(dp, q);
			for(int j = q-1;j >= 0;j--){
				int inf = j+(a[i]>>>6);
				if(inf+1 < q && (a[i]&63)>0)dp[inf+1] |= dp[j]>>>-a[i];
				if(inf < q)dp[inf] |= dp[j]<<a[i];
			}
			for(int j = 0;j < q;j++){
				for(long f = dp[j]^pre[j];f != 0L;f &= f-1){
					last[j<<6|Long.numberOfTrailingZeros(f)] = i;
				}
			}
		}
		if(dp[ss>>>6]<<63-ss<0){
			int[][] b = new int[n][];
			int p = 0;
			int cur = ss;
			boolean[] used = new boolean[n];
			while(cur > 0){
				used[last[cur]] = true;
				cur -= a[last[cur]];
			}
			for(int i = 0;i < n;i++){
				if(!used[i]){
					b[p++] = new int[]{i, a[i]};
				}
			}
			Arrays.sort(b, 0, p, new Comparator<int[]>() {
				public int compare(int[] a, int[] b) {
					return a[1] - b[1];
				}
			});
			int[] rg = new int[n];
			int[] ret = new int[n];
			Arrays.fill(ret, -1);
			Arrays.fill(rg, -1);
			for(int i = p-1;i >= 1;i--){
				ret[b[i][0]] = b[i-1][0];
				rg[b[i][0]] = b[i][1]-b[i-1][1];
			}
			ret[b[0][0]] = -1;
			rg[b[0][0]] = b[0][1];
			for(int i = 0;i < n;i++){
				if(rg[i] == -1)rg[i] = a[i];
				out.print(rg[i] + " ");
				if(ret[i] == -1){
					out.println(0);
				}else{
					out.println(1 + " " + (ret[i]+1));
				}
			}
		}else{
			out.println(-1);
		}
	}
	
	void run() throws Exception
	{
//		int n = 70000, m = 99999;
//		Random gen = new Random();
//		StringBuilder sb = new StringBuilder();
//		sb.append(n + " ");
//		sb.append(n + " ");
//		for(int i = 0;i < n;i++){
//			sb.append(gen.nextInt(70000)+1 + " ");
//		}
//		INPUT = sb.toString();
		
		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 D().run(); }
	
	private byte[] inbuf = new byte[1024];
	private 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)); }
}