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;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vb = vector<bool>;
using pii = pair<int, int>;
using mii = map<int, int>;
using seti = set<int>;
using vpii = vector<pair<int, int>>;
using pqmx = priority_queue<int>;
using pqmn = priority_queue<int, vector<int>, greater<int>>;
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
inline T lcm(T a, T b) {
  return (a / gcd(a, b)) * b;
}
template <typename T>
inline int sz(T &&x) {
  return (int)(x).size();
}
template <typename T, typename t>
inline T minm(T &x, t y) {
  if (y < x) x = y;
  return x;
}
template <typename T, typename t>
inline T maxm(T &x, t y) {
  if (y > x) x = y;
  return x;
}
void Main() {
  int n, k;
  cin >> n >> k;
  vi a(n);
  for (int _ = 0; _ < (n); ++_) cin >> a[_];
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (n); ++j)
      if (abs(a[i] - a[j]) > k) {
        cout << "NO\n";
        return;
      }
  cout << "YES\n";
  for (auto &i : (a)) {
    int c = -1;
    for (int j = (0); j < (i); ++j) cout << (++c) % k + 1 << " ";
    cout << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  for (int t = 1; t <= T; ++t) {
    Main();
  }
  return 0;
}
