There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track.

 

The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has coaches numbered in increasing order . The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be . Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.

Input 

The last block consists of just one line containing 0.

Output 

Sample Input 

5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0

Sample Output 

Yes
No

Yes

有n辆火车  按1到n的顺序进站  最后进站的车能够在不论什么时候出去  推断给定的出站序列是否可能

火车仅仅有两种状态  从A进站  或者从站到B   模拟栈的操作即可了

令A表示A中当前待进站的第一辆火车  tar[B]表示出站序列中当前应该出站的火车   sta为火车站

当A==tar[B]的时候  A进站立即出战   否则当站中最后一辆==tar[B]时   这辆车出站  都不满足就仅仅能A中的最前面的火车进站

当n辆火车所有进站  而站中还有火车是   给定的出战序列就是不可能的

#include<cstdio>
#include<stack>
using namespace std;
const int N = 1005;
int n, tar[N], A, B;
int main()
{
    while (scanf ("%d", &n), n)
    {
        while (scanf ("%d", &tar[1]), tar[1])
        {
            for (int i = 2; i <= n; ++i)
                scanf ("%d", &tar[i]);
            stack<int> sta;
            A = B = 1;
            bool ok = true;
            while (B <= n)
            {
                if (A == tar[B])
                {   ++A; ++B;   }
                else if (!sta.empty() && sta.top() == tar[B])
                {   sta.pop();  ++B;  }
                else if (A <= n)
                    sta.push (A++);
                else
                {   ok = false;  break;  }
            }
            printf (ok ? "Yes\n" : "No\n");
        }
        printf("\n");
    }
    return 0;
}