Problem: 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

Solution (in PYTHON3):

Solution:
# Aluno: Rafael Pontes

n, k = [int(tamanho_pilha) for tamanho_pilha in input().split()]
tamanhos_pilhas = [int(tamanho_pilha) for tamanho_pilha in input().split()]

if k < 1:
    print("NO")
    exit(0)

menor_pilha = 101
for tamanho_pilha in tamanhos_pilhas:
    menor_pilha = min(menor_pilha, tamanho_pilha)

solucao = []
for i in range(n):
    solucao.append(list())
    for j in range(101):
        solucao[i].append(1)

menor_pilha += 1
for i in range(n):
    prox_cor = 2
    for j in range(menor_pilha, tamanhos_pilhas[i]):
        if prox_cor > k:
            print("NO")
            exit(0)
        solucao[i][j] = prox_cor
        prox_cor += 1

print("YES")
for i in range(n):
    for j in range(tamanhos_pilhas[i]):
        print("%d " % solucao[i][j], end="")
    print("")