Problem: Fillomino is a classic logic puzzle. (You do not need to know Fillomino in order to solve this problem.) In one classroom in Yunqi town, some volunteers are playing a board game variant of it:

Consider an n by n chessboard. Its rows are numbered from 1 to n from the top to the bottom. Its columns are numbered from 1 to n from the left to the right. A cell on an intersection of x-th row and y-th column is denoted (x, y). The main diagonal of the chessboard is cells (x, x) for all 1 ≤ x ≤ n.

A permutation of \{1, 2, 3, ..., n\} is written on the main diagonal of the chessboard. There is exactly one number written on each of the cells. The problem is to partition the cells under and on the main diagonal (there are exactly 1+2+ … +n such cells) into n connected regions satisfying the following constraints:

  1. Every region should be connected. That means that we can move from any cell of a region to any other cell of the same region visiting only cells of the same region and moving from a cell to an adjacent cell. 
  2. The x-th region should contain cell on the main diagonal with number x for all 1≤ x≤ n. 
  3. The number of cells that belong to the x-th region should be equal to x for all 1≤ x≤ n. 
  4. Each cell under and on the main diagonal should belong to exactly one region. 

Input

The first line contains a single integer n (1≤ n ≤ 500) denoting the size of the chessboard.

The second line contains n integers p_1, p_2, ..., p_n. p_i is the number written on cell (i, i). It is guaranteed that each integer from \{1, …, n\} appears exactly once in p_1, ..., p_n.

Output

If no solution exists, output -1.

Otherwise, output n lines. The i-th line should contain i numbers. The j-th number on the i-th line should be x if cell (i, j) belongs to the the region with x cells.

Examples

Input


3
2 3 1


Output


2
2 3
3 3 1


Input


5
1 2 3 4 5


Output


1
2 2
3 3 3
4 4 4 4
5 5 5 5 5

Note

The solutions to the examples are illustrated in the following pictures: <image> <image>

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <bitset>
#include <cmath>
#include <string>
#include <stack>
#include <set>
#include <map>
#include <queue>
using namespace std;
template<typename A,typename B>
string to_string(pair<A,B> p);

template<typename A,typename B,typename C>
string to_string(tuple<A,B,C> p);

template<typename A,typename B,typename C,typename D>
string to_string(tuple<A,B,C,D> p);

string to_string(const string &s)
{
    return '"'+s+'"';
}
string to_string(const char *s)
{
    return to_string((string)s);
}
string to_string(bool b)
{
    return (b?"true":"false");
}
string to_string(vector<bool> v)
{
    bool first=true;
    string res="{";
    for(int i=0;i<static_cast<int>(v.size());i++)
    {
        if(!first)
        {
            res+=", ";
        }
        first=false;
        res+=to_string(v[i]);
    }
    res+="}";
    return res;
}
template<size_t N>
string to_string(bitset<N> v)
{
    string res="";
    for(size_t i=0;i<N;i++)
    {
        res+=static_cast<char>('0'+v[i]);
    }
    return res;
}
template <typename A>
string to_string(A v)
{
    bool first=true;
    string res="{";
    for(const auto &x:v)
    {
        if(!first)
        {
            res+=", ";
        }
        first=false;
        res+=to_string(x);
    }
    res+="}";
    return res;
}
template<typename A,typename B>
string to_string(pair<A, B> p)
{
    return "("+to_string(p.first)+", "+to_string(p.second)+")";
}
template <typename A,typename B,typename C>
string to_string(tuple<A,B,C> p)
{
    return "("+to_string(get<0>(p))+", "+to_string(get<1>(p))+", "+to_string(get<2>(p))+")";
}
template<typename A,typename B,typename C,typename D>
string to_string(tuple<A,B,C,D> p)
{
    return "("+to_string(get<0>(p))+", "+to_string(get<1>(p))+","+to_string(get<2>(p))+", "+to_string(get<3>(p))+")";
}
void debug(){cerr<<endl;}
template<typename Head, typename... Tail>
void debug(Head H,Tail...T)
{
    cerr<<" "<<to_string(H);
    debug(T...);
}
struct Fast_power
{
    long long integer(long long n,long long e,long long my_mod=9223372036854775807LL)
    {
        long long ans=1;
        while(e)
        {
            if(e&1) ans=(ans*n)%my_mod,e--;
            else n=(n*n)%my_mod,e>>=1;
        }
        return ans;
    }
    vector<vector<long long>> matrix_multiply(vector<vector<long long>> a,vector<vector<long long>> b)
    {
        vector<vector<long long>> c(a.size(),vector<long long>(b[0].size(),0));
        for(int i=0;i<a.size();i++)
            for(int q=0;q<b[0].size();q++)
            {
                long long num=0;
                for(int e=0;e<a[0].size();i++)
                    num+=a[i][e]*b[e][q];
                c[i][q]=num;
            }
        return c;
    }
    // long long matrix(vector<vector<long long>> &n,long long e,vector<long long> &s,long long my_mod=9223372036854775807LL)//矩阵快速幂
    // {
    //     vector<vector<long long>> ans(n.size(),vector<long long>(n.size(),1));
    //     while(e)
    //     {
    //         if(e&1) ans=matrix_multiply(ans,n),e--;
    //         else n=matrix_multiply(n,n),e>>=1;
    //     }
    // }
};
struct Euclidean//欧几里得 
{
    int gcdi(int a,int b)
    {
        return b?gcdi(b,a%b):a;
    }
    double gcdf(double a,double b,double accuracy)//浮点数的最大公约数
    {
        return b<=accuracy?a:gcdf(b,fmod(a,b),accuracy);
    }
    //扩展欧几里得

    //类欧几里得
};

#define DijkstraDist long long
const int maxnD=100000+5;
const DijkstraDist INFD=0x3f3f3f3f;;
struct Dijkstra
{
    struct Edge
    {
        int from,to;
        DijkstraDist dist;
    };
    struct HeapNode
    {
        DijkstraDist d;
        int u;
        bool operator < (const HeapNode&rhs) const
        {
            return d>rhs.d;
        }
    };
    vector<Edge> edges;
    vector<int> G[maxnD];
    bool done[maxnD];
    DijkstraDist d[maxnD];
    int p[maxnD],n;
    void init(int n)
    {
        this->n=n;
        for(int i=0;i<=n;i++) G[i].clear();
        edges.clear();
    }
    void AddEdge(int from,int to,DijkstraDist dist)
    {
        edges.push_back((Edge){from,to,dist});
        int m=edges.size();
        G[from].push_back(m-1);
    }
    void dijkstra(int s)
    {
        priority_queue<HeapNode> q;
        memset(d,INFD,sizeof(d));
        d[s]=0;
        memset(done,0,sizeof(done));
        q.push((HeapNode){0,s});
        while(!q.empty())
        {
            HeapNode x=q.top();
            q.pop();
            int u=x.u;
            if(done[u]) continue;
            done[u]=true;
            for(int i=0;i<G[u].size();i++)
            {
                Edge& e=edges[G[u][i]];
                if(d[e.to]>d[u]+e.dist)
                {
                    d[e.to]=d[u]+e.dist;
                    p[e.to]=G[u][i];
                    q.push((HeapNode){d[e.to],e.to});
                }
            }
        }
    }
};

const int maxnL=500000+5;
const int INFL=0x3f3f3f3f;
struct LIS
{
    int g[maxnL],a[maxnL],d[maxnL];
    void lis(int n)
    {
        for(int i=1;i<=n;i++) g[i]=INFL;
        for(int i=1;i<n;i++)
        {
            int k=lower_bound(g+1,g+n+1,a[i])-g;
            g[k]=a[i];
        }
    }
};
const int maxnS=200000+5;
struct SegmentTree
{
    int a[maxnS];
    int sumv[maxnS*4],minv[maxnS*4],maxv[maxnS*4],addv[maxnS*4],setv[maxnS*4];
    int lc,rc,v;
    int _min,_max,_sum;
    void build(int l,int r,int o)
    {
        if(l==r) return void(sumv[o]=minv[o]=maxv[o]=a[l]);
        build(l,(l+r)>>1,o<<1);
        build(((l+r)>>1)+1,r,o<<1|1);
    }
    void maintain(int l,int r,int o)
    {
        sumv[o]=minv[o]=maxv[o]=a[l];
        if(r>l)
        {
            sumv[o]=sumv[o<<1]+sumv[o<<1|1];
            minv[o]=min(minv[o<<1],minv[o<<1|1]);
            maxv[o]=max(maxv[o<<1],maxv[o<<1|1]);
        }
        minv[o]+=addv[o],maxv[o]+=addv[o],sumv[o]+=addv[o]*(r-l+1);
    }
    void update(int l,int r,int o)
    {
        if(l>=lc&&r<=rc) addv[o]+=v;
        else
        {
            int m=(l+r)>>1;
            if(lc<=m) update(l,(l+r)>>1,o<<1);
            if(rc>m) update(((l+r)>>1)+1,r,o<<1|1);
        }
        maintain(l,r,o);
    }
    void query(int l,int r,int o,int add)
    {
        if(lc<=l&&r<=rc) 
        {
            _sum+=sumv[o]+add*(r-l+1);
            _min=min(_min,minv[o]+add);
            _max=max(_max,maxv[o]+add);
        }
        else
        {
            int m=(l+r)>>1;
            if(lc<=m) query(l,m,o<<1,add+addv[o]);
            if(rc>m) query(m+1,r,o<<1|1,add+addv[o]);
        }
    }
};

const int maxnB=400000+5;
struct BIT
{
    int c[maxnB],n; 
    int lowbit(int x)
    {
        return x&-x;
    }
    int sum(int x)
    {
        int ret=0;
        while(x>0)
        {
            ret+=c[x];
            x-=lowbit(x);
        }
        return ret;
    }
    void add(int x,int d)
    {
        while(x<=n)
        {
            c[x]+=d;
            x+=lowbit(x);
        }
    }
};

const int maxnEC=200000+5;
struct EXCRT//扩展中国剩余定理
{
    //x%b[i]=a[i];
    long long ai[maxnEC],bi[maxnEC],n;
    inline long long exgcd(long long a,long long b,long long &x,long long &y)
    {
        if(!b)
        {
            x=1,y=0;
            return a;
        }
        long long gcd=exgcd(b,a%b,x,y);
        long long t=x;
        x=y,y=t-(a/b)*y;//递归解方程。 
        return gcd;
    }
    inline long long mul(long long a,long long b,long long mod) //必须自己打乘法不然会爆long long 
    {
        long long s=0;
        while(b) 
        {
            if(b&1) s=(s+a)%mod;
            a=(a+a)%mod;
            b>>=1;
        }
        return s%mod;
    }
    long long excrt()
    {
        long long x=0,y=0,k;
        long long M=bi[1],ans=ai[1];//第一个方程的解特判
        for(int i=2;i<=2;i++)
        {
            long long a=M,b=bi[i],c=(ai[i]-ans%b+b)%b;//ax≡c(mod b)
            long long gcd=exgcd(a,b,x,y),bg=b/gcd;
            if(c%gcd!=0) return -1; //是否无解
            x=mul(x,c/gcd,bg);
            ans+=x*M;//更新前k个方程组的答案
            M*=bg;//M为前k个m的lcm
            ans=(ans%M+M)%M;
        }
        return (ans%M+M)%M;
    }
};

#define PI acos(-1);
#define fri freopen("in.txt", "r", stdin)
#define fro freopen("out.txt", "w", stdout)
#define pb push_back
#define all(a) a.begin(),a.end()
#define si(a) a.size()
#define in insert
#define fora(i,l,r) for(int i=l;i<r;i++)
#define forb(i,l,r) for(int i=l;i<=r;i++)
#define forc(i,l,r) for(int i=l;i>r;i--)
#define ford(i,l,r) for(int i=l;i>=r;i--)
#define fi first
#define ed second
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<long long> vl;
typedef pair<int,int> pi;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=500+5;
int vis[maxn][maxn];
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    int a[maxn];
    cin>>n;
    fora(i,0,n) cin>>a[i];
    int ok=1;
    fora(i,0,n)
    {
        int num=a[i];
        a[i]--;
        vis[i][i]=num;
        int fx=i,fy=i;
        while(a[i]--)
        {
            if(fy>0&&!vis[fx][fy-1])
            {
                fy--;
                vis[fx][fy]=num;
                continue;
            }
            if(fx<n-1&&!vis[fx+1][fy])
            {
                fx++;
                vis[fx][fy]=num;
                continue;
            }
            {ok=0;break;}
        }
    }
    if(ok)
    {
        fora(i,0,n)
        {
            fora(q,0,i+1) cout<<vis[i][q]<<" ";
            cout<<endl;
        }
    }
    else cout<<-1<<endl;
	return 0;
}