There is a bag-like data structure, supporting two operations:

1 x

Throw an element x into the bag.

2

Take out an element from the bag.

Given a sequence of operations with return values, you‘re going to guess the data structure. It is a stack (Last-In, First-Out), a queue (First-In, First-Out), a priority-queue (Always take out larger elements first) or something else that you can hardly imagine!

Input

There are several test cases. Each test case begins with a line containing a single integer n (1<=n<=1000). Each of the next n lines is either a type-1 command, or an integer 2 followed by an integer x. That means after executing a type-2 command, we get an element x without error. The value of x is always a positive integer not larger than 100. The input is terminated by end-of-file (EOF). The size of input file does not exceed 1MB.

Output

For each test case, output one of the following:

stack

It‘s definitely a stack.

queue

It‘s definitely a queue.

priority queue

It‘s definitely a priority queue.

impossible

It can‘t be a stack, a queue or a priority queue.

not sure

It can be more than one of the three data structures mentioned above.

Sample Input

6
1 1
1 2
1 3
2 1
2 2
2 3
6
1 1
1 2
1 3
2 3
2 2
2 1
2
1 1
2 2
4
1 2
1 1
2 1
2 2
7
1 2
1 5
1 1
1 3
2 5
1 4
2 4

Output for the Sample Input

queue
not sure
impossible
stack
priority queue

题意：有一个类似”包“的数据结构，支持两种操作：

1 a   把元素a放进包里

2 a   从包里取出a

给出一系列操作，问你这个包代表的是什么.可能是栈，可能是队列，也有可能是优先队列（数值大的先出）或者其他的东西。

分析：题目不难，要注意一个问题：当”包“为空时，不能从包里取东西。

代码：

#include<stdio.h>
#include<stack>
#include<queue>
using namespace std;
int main()
{
    int n, i, k, a, f[4];
    while(~scanf("%d",&n))
    {
        stack<int> s;
        queue<int> q;
        priority_queue<int, vector<int>, less<int> > pq;
        for(i = 0; i < 3; i++) f[i] = 1;
        for(i = 0; i < n; i++)
        {
            scanf("%d%d",&k,&a);
            if(k == 1)
            {
                s.push(a);
                q.push(a);
                pq.push(a);
            }
            else
            {
                if(!s.empty())
                {
                    if(s.top() != a)
                        f[0] = 0;
                    s.pop();
                }
                else
                    f[0] = 0;

                if(!q.empty())
                {
                    if(q.front() != a)
                        f[1] = 0;
                    q.pop();
                }
                else
                    f[1] = 0;

                if(!pq.empty())
                {
                    if(pq.top() != a)
                        f[2] = 0;
                    pq.pop();
                }
                else
                    f[2] = 0;
            }
        }
        int cnt = 0;
        for(i = 0; i < 3; i++)
            if(f[i] == 1)
                cnt++;
        if(cnt == 0)
            printf("impossible\n");
        else if(cnt > 1)
            printf("not sure\n");
        else
        {
            if(f[0] == 1)
                printf("stack\n");
            else if(f[1] == 1)
                printf("queue\n");
            else
                printf("priority queue\n");
        }
    }
    return 0;
}

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