Problem: Dima's spent much time thinking what present to give to Inna and gave her an empty sequence w. Now they want to fill sequence w with numbers zero and one. For that, they decided to play an amusing game. 

Before the game begins, Dima chooses m integers a1, a2, ..., am (1 ≤ a1 < a2 < ... < am). Then Inna and Dima start playing, that is, adding numbers to sequence w. Each new number they choose is added to the end of the sequence. At some moments of time Dima feels that the game is going to end too soon (and he wants to play with Inna as long as possible), so he hits a table hard with his fist. At that the a1-th, a2-th, a3-th, ..., ak-th numbers from the beginning simultaneously fall out of the sequence (the sequence gets k numbers less). Here k is such maximum number that value ak doesn't exceed the current length of the sequence. If number a1 is larger than the current length of w, then nothing falls out of the sequence.

You are given the chronological sequence of events in the game. Each event is either adding a number to the end of sequence w or Dima's hit on the table. Calculate the sequence w after all these events happen.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 106) showing how many events took place and how many numbers Dima chose.

The next line contains m distinct integers ai (1 ≤ ai ≤ 106) sorted in the increasing order. 

Next n lines describe the events in the chronological order. Each line contains a single integer: -1, 0 or 1. Number -1 means that Dima hits the table. Number 0 means that Inna and Dima add number 0 to the end of the sequence. Number 1 means that Inna and Dima add number 1 to the end of the sequence.

Output

In a single line print a sequence of numbers 0 and 1 — the elements of the sequence after all events happen. Print the elements of the sequence in the order from the beginning to the end of the sequence.

If after all events the sequence ends up empty, print "Poor stack!".

Examples

Input

10 3
1 3 6
-1
1
1
0
0
-1
0
1
-1
1


Output

011


Input

2 1
1
1
-1


Output

Poor stack!

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Map.Entry;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

import javax.swing.event.TreeSelectionEvent;

public class InnaAndSequence_R220D {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[] arr = new int[m];
		TreeMap<Integer, Integer> map = new TreeMap<>();
		for (int i = 0; i < m; i++) {
			arr[i] = sc.nextInt();
			map.put(arr[i], i);
		}
		FenwickTree ft = new FenwickTree(n);
		ArrayList<Integer> nums = new ArrayList<>();
		int cnt = 0;
		int idxx = 0;
		for (int i = 1; i <= n; i++) {
			int x = sc.nextInt();
			if (x == -1) {
				Entry<Integer, Integer> en = map.floorEntry(cnt);
				if (en != null) {
					int st = en.getValue();
					for (int j = st; j >= 0; j--) {
						int now = arr[j];
						cnt--;
						int lo = 1, hi = n, ans = -1;
						while (lo <= hi) {
							int mid = ((lo + hi) >> 1);
							if (ft.rsq(mid) >= now) {
								hi = mid - 1;
								ans = mid;
							} else
								lo = mid + 1;
						}
						int idx = ans;
						ft.point_update(idx, -1);
					}
				}
			} else {
				nums.add(x);
				cnt++;
				ft.point_update(idxx + 1, 1);
				++idxx;
			}
		}
		PrintWriter pw = new PrintWriter(System.out);
		boolean ok = false;
		for (int i = 1; i <= n; i++) {
			if (ft.point_query(i) == 1) {
				ok = true;
				pw.print(nums.get(i - 1));
			}
		}
		if (!ok)
			pw.println("Poor stack!");
		else
			pw.println();
		pw.flush();
	}

	static class FenwickTree { // one-based DS

		int n;
		int[] ft;

		FenwickTree(int size) {
			n = size;
			ft = new int[n + 1];
		}

		int rsq(int b) // O(log n)
		{
			int sum = 0;
			while (b > 0) {
				sum += ft[b];
				b -= b & -b;
			} // min?
			return sum;
		}

		int rsq(int a, int b) {
			return rsq(b) - rsq(a - 1);
		}

		void point_update(int k, int val) // O(log n), update = increment
		{
			while (k <= n) {
				ft[k] += val;
				k += k & -k;
			} // min?
		}

		int point_query(int idx) // c * O(log n), c < 1
		{
			int sum = ft[idx];
			if (idx > 0) {
				int z = idx ^ (idx & -idx);
				--idx;
				while (idx != z) {
					sum -= ft[idx];
					idx ^= idx & -idx;
				}
			}
			return sum;
		}

		void scale(int c) {
			for (int i = 1; i <= n; ++i)
				ft[i] *= c;
		}

		int findIndex(int cumFreq) {
			int tmp = cumFreq;
			int msk = n;
			while ((msk & (msk - 1)) != 0)
				msk ^= msk & -msk; // msk will contain the MSB of n

			int idx = 0;
			while (msk != 0) {
				int tIdx = idx + msk;
				if (tIdx <= n && cumFreq > ft[tIdx]) {
					idx = tIdx;
					cumFreq -= ft[tIdx];
				}
				msk >>= 1;
			}
			// if (cumFreq != 0)
			// return -1;
			return idx + 1 > n ? -1 : ft[idx + 1] == cumFreq ? idx + 1 : -1;
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}
	}
}