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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define f first
#define s second
using namespace std;
using ll = long long;
using ii = pair<int, int>;
constexpr int MN = 505;

int ans[MN*MN];
ii x[MN];

int main() {
    if (fopen("in", "r")) freopen("in", "r", stdin), freopen("out", "w", stdout);
    ios_base::sync_with_stdio(0), cin.tie(0);

    int N; cin >> N;
    set<int> S;
    for (int i = 0; i < N*N; ++i) S.insert(i);
    for (int i = 0; i < N; ++i) {
        cin >> x[i].f;
        S.erase(--x[i].f), x[i].s = i;
    }
    sort(x, x+N);    
    for (int i = 0; i < N; ++i) {
        int j = 0;
        while (S.size() && *begin(S) < x[i].f && j < x[i].s) {
            ans[*begin(S)] = x[i].s;
            S.erase(begin(S));
            ++j;
        }
        if (j < x[i].s) {
            cout << "No";
            return 0;
        }
        ans[x[i].f] = x[i].s;
    }
    for (int i = 0; i < N; ++i) {
        int j = 0;
        while (S.size() && j < N-i-1) {
            ans[*begin(S)] = i;
            S.erase(begin(S));
            ++j;
        }
    }
    cout << "Yes\n";
    for (int i = 0; i < N*N; ++i) cout << ans[i]+1 << ' ';
}