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 

Solution：

方法很簡單，每次找比a[i]小的第一個數，然後插到尾部就行，沒找到的話另起一行。暴力找複雜度太高，又因為尾部的數從上往下是嚴格遞減的，所以需要把尾部的數存到一個數組中，用二分的方法查詢，能把時間縮短到O(logn)。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <map>
#include <queue>
#include <stack>

using namespace std;

typedef long long ll;

const int maxn = 1e5 + 100000;
int a[maxn];
int temp[maxn];
vector < int > ans[maxn];

int Bin(int l, int r, int x)
{
    int res = -1;
    while(l <= r)
    {
        int mid = (l + r) >> 1;
        if(temp[mid] > x)
        {
            l = mid + 1;
        }
        else
        {
             res = mid;
             r = mid - 1;
        }
    }
    return res;
}

int main()
{
    int n;
    cin >> n;
    for(int i = 1; i <= n; ++ i)
    {
        scanf("%d", &a[i]);
    }
    int top = 0;
    for(int i = 1; i <= n; ++ i)
    {
        int cnt = Bin(0, top - 1, a[i]);
        if(cnt == -1)
        {
            ans[top].push_back(a[i]);
            temp[top] = a[i];
            top++;
        }
        else
        {
            ans[cnt].push_back(a[i]);
            temp[cnt] = a[i];
        }
    }
    for(int i = 0; i < top; ++ i)
    {
        for(int j = 0; j < ans[i].size(); ++ j)
        {
            if(j == 0)
                cout << ans[i][j];
            else
                cout << ' ' << ans[i][j];
        }
        cout << endl;
    }
    return 0;
}