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;
const int N = 505;
int n, sum, pt, num[N], p[N], ans[N * N], ex[N];
struct pr {
   	int x, i;
} a[N];
bool cmp (pr u, pr v) {return u.x < v.x;} 
struct pri {
	int i, x;
	pri () {}
	pri (int _i, int _x) :
		i(_i), x(_x) {}
	bool operator < (const pri &o) const {
		return p[i] - (i - num[i]) > p[o.i] - (o.i - num[o.i]);
	}
};
priority_queue <pri> q;
int main () {
	cin >> n, sum = 0;
	for (int i = 1; i <= n; ++i)
		cin >> a[i].x, p[i] = a[i].x, a[i].i = i, ans[a[i].x] = i;
	sort(a + 1, a + 1 + n, cmp);
	pt = n * n + 1;
	for (int i = 1; i <= n; ++i) if (a[i].i != 1) q.push(pri(a[i].i, a[i].x));
	for (int i = 1; i <= n * n; ++i) {
		if (!ans[i]) {
			if (q.empty()) {pt = i; break;}
			pri t = q.top(); q.pop();
			while (p[t.i] < i && !q.empty()) t = q.top(), q.pop();
			if (p[t.i] < i && q.empty()) break;
			ans[i] = t.i, ++num[t.i];
			if (p[t.i] < i || num[t.i] == t.i - 1) continue;
			q.push(t);
		} else ++num[ans[i]];
	}
	for (int i = pt; i <= n * n; ++i) {
		if (ans[i]) {++num[ans[i]]; continue;}
		for (int j = 1; j <= n; ++j) {
			if (num[j] < n && p[j] < i) {ans[i] = j; ++num[j]; break;}
			if (num[j] < n - 1) {ans[i] = j; ++num[j]; break;}
		}
	}
	for (int i = 1; i <= n * n; ++i) ++ex[ans[i]];
	for (int i = 1; i <= n; ++i) if (ex[i] != n) return puts("No"), 0;
	for (int i = 1; i <= n; ++i) if (ans[p[i]] != i) return puts("No"), 0;
	for (int i = 1; i <= n; ++i) num[i] = 0;
	for (int i = 1; i <= n * n; ++i) {
		++num[ans[i]];
		if (i == p[ans[i]] && num[ans[i]] != ans[i]) {return puts("No"), 0;}
	}
	puts("Yes");
	for (int i = 1; i <= n * n; ++i)
		printf("%d%c", ans[i], i < n * n ? ' ' : '\n');
	return 0;
}