Problem:
Score : 800 points

Problem Statement

You are given an integer sequence x of length N.
Determine if there exists an integer sequence a that satisfies all of the following conditions, and if it exists, construct an instance of a.

a is N^2 in length, containing N copies of each of the integers 1, 2, ..., N.

For each 1 ≤ i ≤ N, the i-th occurrence of the integer i from the left in a is the x_i-th element of a from the left.

Constraints

1 ≤ N ≤ 500

1 ≤ x_i ≤ N^2

All x_i are distinct.

Input

The input is given from Standard Input in the following format:

N
x_1 x_2 ... x_N

Output

If there does not exist an integer sequence a that satisfies all the conditions, print No.
If there does exist such an sequence a, print Yes in the first line, then print an instance of a in the second line, with spaces inbetween.

Sample Input 1

3
1 5 9

Sample Output 1

Yes
1 1 1 2 2 2 3 3 3

For example, the second occurrence of the integer 2 from the left in a in the output is the fifth element of a from the left.
Similarly, the condition is satisfied for the integers 1 and 3.

Sample Input 2

2
4 1

Sample Output 2

No

Solution in Ruby (2.3.3) (judge verdict: Runtime Error)

#include<bits/stdc++.h>

#define PB push_back
#define SZ(c) int((c).size())
#define ALL(c) (c).begin(), (c).end()
#define CONTAINS(c, x) ((c).find(x) != (c).end())
#define REP(i, n) for(int i=1; i<=n; i++)
#define WAIT cout<<flush, system("PAUSE");
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int MAX = 300100;

int N;
set<int> S;
int X[MAX], A[MAX];

void NO(){
	cout << "No" << endl;
	exit(0);
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	cin >> N;
	REP(I, N) cin >> X[I];
	REP(I, N*N) S.insert(I);
	
	REP(I, N) {
		if (!CONTAINS(S, X[I])) NO();
		A[ X[I] ] = I;
		S.erase(X[I]);
		auto base = S.lower_bound(X[I]);
		
		REP(K, I-1){
			auto it = base;
			if (it == S.begin()) NO();
			it--;
			A[ *it ] = I;
			S.erase(it);
		}
	}
	REP(I, N) REP(K, N-I){
		A[*S.begin()] = I;
		S.erase(S.begin());
	}
	cout << "Yes" << endl;
	REP(I, N*N) cout << A[I] << " \n"[I==N*N];
	// REP <=
}