Problem: Dima has a birthday soon! It's a big day! Saryozha's present to Dima is that Seryozha won't be in the room and won't disturb Dima and Inna as they celebrate the birthday. Inna's present to Dima is a stack, a queue and a deck.

Inna wants her present to show Dima how great a programmer he is. For that, she is going to give Dima commands one by one. There are two types of commands:

  1. Add a given number into one of containers. For the queue and the stack, you can add elements only to the end. For the deck, you can add elements to the beginning and to the end. 
  2. Extract a number from each of at most three distinct containers. Tell all extracted numbers to Inna and then empty all containers. In the queue container you can extract numbers only from the beginning. In the stack container you can extract numbers only from the end. In the deck number you can extract numbers from the beginning and from the end. You cannot extract numbers from empty containers. 



Every time Dima makes a command of the second type, Inna kisses Dima some (possibly zero) number of times. Dima knows Inna perfectly well, he is sure that this number equals the sum of numbers he extracts from containers during this operation.

As we've said before, Dima knows Inna perfectly well and he knows which commands Inna will give to Dima and the order of the commands. Help Dima find the strategy that lets him give as more kisses as possible for his birthday!

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of Inna's commands. Then n lines follow, describing Inna's commands. Each line consists an integer:

  1. Integer a (1 ≤ a ≤ 105) means that Inna gives Dima a command to add number a into one of containers. 
  2. Integer 0 shows that Inna asks Dima to make at most three extractions from different containers. 

Output

Each command of the input must correspond to one line of the output — Dima's action.

For the command of the first type (adding) print one word that corresponds to Dima's choice:

  * pushStack — add to the end of the stack; 
  * pushQueue — add to the end of the queue; 
  * pushFront — add to the beginning of the deck; 
  * pushBack — add to the end of the deck. 



For a command of the second type first print an integer k (0 ≤ k ≤ 3), that shows the number of extract operations, then print k words separated by space. The words can be:

  * popStack — extract from the end of the stack; 
  * popQueue — extract from the beginning of the line; 
  * popFront — extract from the beginning from the deck; 
  * popBack — extract from the end of the deck. 



The printed operations mustn't extract numbers from empty containers. Also, they must extract numbers from distinct containers.

The printed sequence of actions must lead to the maximum number of kisses. If there are multiple sequences of actions leading to the maximum number of kisses, you are allowed to print any of them.

Examples

Input

10
0
1
0
1
2
0
1
2
3
0


Output

0
pushStack
1 popStack
pushStack
pushQueue
2 popStack popQueue
pushStack
pushQueue
pushFront
3 popStack popQueue popFront


Input

4
1
2
3
0


Output

pushStack
pushQueue
pushFront
3 popStack popQueue popFront

Solution (in JAVA):

Solution:
//package a2oj_ladder_C;

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;


@SuppressWarnings("unused")
public class Dima358C {

	InputStream is;
	PrintWriter out;
	String INPUT = "";

	int mod = (int)(Math.pow(10, 9)+7);

	void solve()
	{	
		Deque<Integer> deque = new LinkedList<>();
		Stack<Integer> st = new Stack<>();
		Queue<Integer> q = new LinkedList<>();
		int n = ni();
		int a[] = new int[n];
		for(int i = 0 ; i<n ; i++) a[i] = ni();

		int c[] = new int[n];
		String s[] = new String[n];

		Arrays.fill(s, "pushFront");

		for(int i = 0 ; i<n ; ) {
			int max1 = -1 , max2 = -1 , max3 = -1;
			int k = 0;
			int p1 = -1 , p2 = -1 , p3 = -1; 
			int j = i;
			while(j<n && a[j] != 0) {

				if(a[j]>=max1) {
					max3 = max2;
					p3 = p2;

					max2 = max1;
					p2 = p1;

					max1 = a[j];
					p1 = j;
				}
				else if(a[j] >= max2) {
					max3 = max2;
					p3 = p2;

					max2 = a[j];
					p2 = j;
				}
				else if(a[j] > max3) {
					max3 = a[j];
					p3 = j;
				}

				j++;
			}
			if(p1 != -1) {
				s[p1] = "pushBack";
			}
			if(p2 != -1) {
				s[p2] = "pushQueue";
			}
			if(p3 != -1) {
				s[p3] = "pushStack";
			}

			if(j<n && a[j] == 0) {
				if(p1 != -1 && p2 != -1 && p3 != -1) {
					s[j] = "3 popBack popQueue popStack";
				}
				else if(p1 != -1 && p2 != -1) {
					s[j] = "2 popBack popQueue";
				}
				else if(p1 != -1) {
					s[j] = "1 popBack";
				}
				else s[j] = "0";
			}

			i = j+1;
		}

		for(String i : s) {
			out.println(i);
		}
	}


	void run() throws Exception
	{
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
	}

	public static void main(String[] args) throws Exception { new Dima358C().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[][] na(int n , int m)
	{
		int[][] a = new int[n][m];
		for(int i = 0;i < n;i++) 
			for(int j = 0 ; j<m ; j++) a[i][j] = 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();
		}
	}

	void display2D(int a[][]) {
		for(int i[] : a) {
			for(int j : i) {
				out.print(j+" ");
			}
			out.println();
		}
	}

	int[][] creategraph(int n , int m) {
		int g[][] = new int[n+1][];

		int from[] = new int[m];
		int to[] = new int[m];
		int ct[] = new int[n+1];

		for(int i = 0 ; i<m; i++) {
			from[i] = ni();
			to[i] = ni();
			ct[from[i]]++;
			ct[to[i]]++;
		}

		int parent[] = new int[n+1];

		for(int i = 0 ; i<n+1 ; i++) g[i] = new int[ct[i]];

		for(int i = 0 ; i<m ; i++) {
			g[from[i]][--ct[from[i]]] = to[i];
			g[to[i]][--ct[to[i]]] = from[i];
		}

		return g;
	}

	static long __gcd(long a, long b)
	{

		if(b == 0) 
		{
			return a;
		}
		else
		{
			return __gcd(b, a % b);
		}
	}

	// To compute x^y under modulo m
	static long power(long x, long y, long p)
	{

		// Initialize result
		long res = 1;

		// Update x if it is more than or
		// equal to p
		x = x % p;

		while (y > 0)
		{

			// If y is odd, multiply x
			// with result
			if (y % 2 == 1)
				res = (res * x) % p;

			// y must be even now
			y = y >> 1; // y = y/2
		x = (x * x) % p;
		}

		return res;
	}

	// Function to find modular 
	// inverse of a under modulo m
	// Assumption: m is prime
	static long modInverse(long a, int m)
	{
		if (__gcd(a, m) != 1) {
			//System.out.print("Inverse doesn't exist");
			return -1;
		}

		else {

			// If a and m are relatively prime, then
			// modulo inverse is a^(m-2) mode m
			//			System.out.println("Modular multiplicative inverse is "
			//											+power(a, m - 2, m));
			return power(a, m - 2, m);
		}
	}

	static long nCrModPFermat(int n, int r,
			int p , long fac[])
	{

		// Base case
		if (r == 0)
			return 1;

		// Fill factorial array so that we
		// can find all factorial of r, n
		// and n-r
		long t = (fac[n]* modInverse(fac[r], p))%p;

		return ( (t* modInverse(fac[n-r], p))% p);
	}


	private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}