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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>

using namespace std;

int N;
vector< pair<int, int> > vec;
vector<int> dust;
int cnt[505];
int ans[250005];

int main(void)
{
	cin >> N;
	int x;
	for(int i = 1; i <= N; i++){
		cin >> x;
		vec.push_back( make_pair(x, i) );
	}
	sort(vec.begin(), vec.end());
	
	int nid = 0, tid = 0, set;
	for(int i = 1; i <= N*N; i++){
		if(nid < vec.size() && i >= vec[nid].first){
			if(cnt[vec[nid].second] < vec[nid].second-1){
				cout << "No" << endl;
				return 0;
			}
			ans[i] = vec[nid].second;
			cnt[vec[nid].second]++;
			for(int j = 0; j < N - vec[nid].second; j++) dust.push_back(vec[nid].second); 
			nid++;
			continue;
		}
		
		while(tid < vec.size() &&cnt[vec[tid].second] >= vec[tid].second-1) tid++;
		if(tid >= vec.size()){
			if(dust.size() == 0){
				cout << "No" << endl;
				return 0;
			}
			set = dust.back();
			dust.pop_back();
		}
		else set = vec[tid].second;
		ans[i] = set;
		cnt[set]++;
	}
	
	cout << "Yes" << endl;
	for(int i = 1; i <= N*N; i++){
		cout << ans[i];
		if(i != N*N) cout << " ";
	}
	cout << endl;
	return 0;
}