Preparing for Merge Sort CodeForces - 847B

Ivan has an array consisting of n different integers. He decided to reorder all elements in increasing order. Ivan loves merge sort so he decided to represent his array with one or several increasing sequences which he then plans to merge into one sorted array.

Ivan represent his array with increasing sequences with help of the following algorithm.

While there is at least one unused number in array Ivan repeats the following procedure:

• iterate through array from the left to the right;
• Ivan only looks at unused numbers on current iteration;
• if current number is the first unused number on this iteration or this number is greater than previous unused number on current iteration, then Ivan marks the number as used and writes it down.

For example, if Ivan's array looks like [1, 3, 2, 5, 4] then he will perform two iterations. On first iteration Ivan will use and write numbers [1, 3, 5], and on second one — [2, 4].

Write a program which helps Ivan and finds representation of the given array with one or several increasing sequences in accordance with algorithm described above.

Input

The first line contains a single integer n (1 ≤ n ≤ 2·105) — the number of elements in Ivan's array.

The second line contains a sequence consisting of distinct integers a1, a2, ..., an (1 ≤ ai ≤ 109) — Ivan's array.

Output

Print representation of the given array in the form of one or more increasing sequences in accordance with the algorithm described above. Each sequence must be printed on a new line.

Examples

Input

5
1 3 2 5 4

Output

1 3 5 
2 4 

Input

4
4 3 2 1

Output

4 
3 
2 
1 

Input

4
10 30 50 101

Output

10 30 50 101 

题意

给你N个不同的数 让你最小行 每行为递增数组

思路

所有行最后一个数是按递减的，我们可以反向存，那么就是递增的，就可以用二分了，开一个数组只存每行的最后一个，二分查找当前数第一个大于他的，把他放在前一行

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 200005;
int a[maxn];
vector<int> v[maxn];
int main()
{
    int n,i,j,m,x;
    scanf("%d",&n);
    memset(a,0,sizeof(a));
    for(i=1;i<=n;i++)
    {
        scanf("%d",&x);
        int it = lower_bound(a,a+n+1,x) - a -1;
        //cout<<it<<endl;
        a[it] = x;
        v[it].push_back(x);
    }
    for(i=n;i>=0;i--)
    {
        int l = v[i].size();
        if(l)
        {
            for(j=0;j<l;j++)
            {
                if(j) printf(" ");
                printf("%d",v[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}