There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.

In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.

Input

The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.

Output

If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .

Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.

Examples

Input

4 4
1 2 3 4


Output

YES
1
1 4
1 2 4
1 2 3 4


Input

5 2
3 2 4 1 3


Output

NO


Input

5 4
3 2 4 3 5


Output

YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const long long mod = 1e9 + 7;
const int alphabet = 26;
const int inf = INT_MAX;
const long long lint = 1e18 + 9;
inline long long cnm(char ch) { return (ch - '0'); }
inline long long cap(char ch) { return (ch - 'a'); }
inline void instr(string& s, long long& n) {
  cin >> s;
  n = s.length();
}
inline void instr(string& s) { cin >> s; }
int a[105];
vector<int> v[105];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, k, m, n;
  cin >> n >> k;
  int mn = inf, mx = 0;
  for (i = 0; i < n; i++) {
    int x;
    cin >> x;
    mx = max(mx, x);
    mn = min(mn, x);
    a[i] = x;
  }
  if ((mx - mn) - k > 0) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << endl;
  for (i = 0; i < n; i++) {
    for (j = 1; j <= mn; j++) v[i].push_back(1);
  }
  long long c = 2;
  for (i = 0; i < n; i++) {
    c = 1;
    for (j = mn + 1; j <= a[i]; j++) {
      v[i].push_back(c++);
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < v[i].size(); j++) {
      cout << v[i][j] << " ";
    }
    cout << endl;
  }
  return 0;
}
