// Original C++ solution (slower):
#define _CRT_SECURE_NO_WARNINGS

#include <iostream>

#include <sstream>

#include <iomanip>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cctype>

#include <cstring>

#include <vector>

#include <list>

#include <queue>

#include <deque>

#include <stack>

#include <map>

#include <set>

#include <algorithm>

#include <iterator>

#include <bitset>

#include <ctime>

#include<complex>

using namespace std;



#define FOR(i,a,b) for (int i = (a); i < (b); i++)

#define RFOR(i,b,a) for (int i = (b)-1; i >= (a); i--)

#define ITER(it,a) for (__typeof(a.begin()) it = a.begin(); it != a.end(); it++)

#define FILL(a,value) memset(a, value, sizeof(a))



#define SZ(a) (int)a.size()

#define ALL(a) a.begin(), a.end()

#define PB push_back

#define MP make_pair



typedef long long LL;

typedef vector<int> VI;

typedef pair<int, int> PII;



const double PI = acos(-1.0);

const int INF = 1000 * 1000 * 1000 + 7;

const LL LINF = INF * (LL)INF;



const int MAX = 1000 * 1000 + 47;

struct Item

{

	int i, pos, left;

	Item(int i1, int pos1, int left1)

	{

		i = i1;

		pos = pos1;

		left = left1;

	}



	Item() {}

};



int A[MAX];

Item X[MAX];



bool cmp(Item a, Item b)

{

	return a.pos < b.pos;

}



void no()

{

	cout << "No" << endl;

	exit(0);

}



int main()

{

	//freopen("in.txt", "r", stdin);

	//freopen("out.txt", "w", stdout);

	ios::sync_with_stdio(false); cin.tie(0);

	int n;

	cin >> n;

	FOR(i, 0, n)

	{

		int x;

		cin >> x;

		x--;

		X[i] = Item(i, x, i);

	}



	sort(X, X + n, cmp);

	FILL(A, -1);

	int p = 0;

	FOR(i, 0, n)

	{

		//cout << X[i].i << " " << X[i].left << endl;

		while (X[i].left != 0)

		{

			if (A[p] == -1)

			{

				X[i].left--;

				A[p] = X[i].i;

			}



			p++;

		}



		if (A[X[i].pos] != -1) no();

		A[X[i].pos] = X[i].i;

	}



	p = 0;

	FOR(i, 0, n * n)

	{

		while (p < n && X[p].left == n - X[p].i - 1) p++;

		if (p == n) break;



		if (A[i] == -1)

		{

			if (X[p].pos > i) no();

			A[i] = X[p].i;

			X[p].left++;

		}

	}



	//FOR(i, 0, n)

	//{

	//	FOR(k, 0, n - i - 1)

	//	{

	//		while (A[p] != -1) p++;

	//		A[p] = X[k].i;

	//	}

	//}



	cout << "Yes" << endl;

	FOR(i, 0, n * n)

	{

		cout << A[i] + 1 << " ";

	}



	cout << endl;

}


// Optimized solution to the same problem (faster):
#include<iostream>

#include<cstring>

#include<cstdio>

#include<algorithm>

#include<utility>

#define gc getchar()

#define fir first

#define sec second

#define N 510

using namespace std;

inline int inn()

{

	int x,ch;while((ch=gc)<'0'||ch>'9');

	x=ch^'0';while((ch=gc)>='0'&&ch<='9')

		x=(x<<1)+(x<<3)+(ch^'0');return x;

}

int fa[N*N],ans[N*N];pair<int,int> x[N];

inline int findf(int x)

{

	int fx=x,y;while(fa[fx]^fx) fx=fa[fx];

	while(x^fx) y=fa[x],fa[x]=fx,x=y;return fx;

}

int main()

{

	int n=inn(),tot=n*n;

	for(int i=1;i<=n;i++) x[i].fir=inn(),x[i].sec=i;

	sort(x+1,x+n+1);for(int i=1;i<=tot;i++) fa[i]=i;

	for(int i=n;i>=1;i--)

	{

		int ps=x[i].fir,t=x[i].sec;ans[ps]=t,fa[ps]=ps-1;

		for(int j=1;j<=n-t;j++)

		{

			int p=findf(tot);

			if(p<ps) return !printf("No\n");

			ans[p]=t,fa[p]=p-1;

		}

	}

	for(int i=n,p;i>=1;i--)

		for(int ps=x[i].fir,t=x[i].sec,j=1;j<t;j++)

			if((p=findf(tot))>ps) return !printf("No\n");

			else ans[p]=t,fa[p]=p-1;

	printf("Yes\n");

	for(int i=1;i<=n*n;i++) printf("%d ",ans[i]);

	return !printf("\n");

}

// Fastest known solution for this problem: