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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define Would
#define you
#define please

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


	int N;
	cin >> N;
	pair<int, int> X[501];
	int kotae[250001];
	rep1(i, N) {
		cin >> X[i].first;
		X[i].second = i;
		kotae[X[i].first] = i;
	}

	sort(X + 1, X + N + 1);

	queue<pair<int, int>> que;
	rep1(i, N) {
		rep(j, X[i].second - 1) que.push(mp(X[i].second, X[i].first));
	}

	int OK = 1;
	rep1(i, N* N) {
		if (kotae[i] == 0) {
			if (que.size() == 0) {
				OK = 0;
				break;
			}
			auto q = que.front();
			que.pop();
			if (q.second < i) {
				OK = 0;
				break;
			}
			kotae[i] = q.first;
		}
		else if (kotae[i] >= 0 && kotae[i] < N) {
			rep(j, N - kotae[i]) que.push(mp(kotae[i], 300000));
		}
	}

	if (OK) {
		co("Yes");
		rep1(i, N * N) co(kotae[i]);
	}
	else co("No");

	Would you please return 0;
}