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:
n,k1=map(int,input().split())
arr=list(map(int,input().split()))
arr1=[]
m=[i for i in range(1,k1+1)]
for i in range(n):
  k=0;j=0;ar=[]
  while k<arr[i]:
    if j==k1:j=0
    ar.append(m[j])
    k+=1;j+=1
  l=[0]*(k1+1)
  for i in ar:l[i]+=1
  arr1.append(l)
flag=True
for i in range(n):
 for j in range(i+1,n):
  for k in range(k1+1):
   if abs(arr1[i][k]-arr1[j][k])>1:flag=False;break
  if flag==False:break
 if flag==False:break
if not flag:print("NO")
else:
  print("YES");ans=[];k1=-1
  for i in arr1:
   ans1=[];k1=-1
   for j in i:
    k1+=1
    ans1+=[k1]*j
   ans.append(ans1)
   ans1=[]
  for i in ans:print(*i)