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
arr=[int(x) for x in input().split()]
n,k=arr[0],arr[1] 
arr=[int(x) for x in input().split()]
for i in range(n):
    arr[i]=[arr[i],i]
arr=sorted(arr,key=lambda x : x[0])    
p1=[0 for i in range(k)]
ans=[[0 for i in range(k)] for i in range(n)]
mini=0
ind=-1
flag=False
#print(ans)
for i in range(n):
    pile=arr[i][0]
    c=pile//k
    m=pile%k
    for j in range(m):
        ans[i][j]+=c+1
       # print(i,j,m,ans,"f1")
        if(j==ind and ans[i][j]>mini+1):
            print("NO")
            flag=True
            break
    for j in range(m,k):
        ans[i][j]=c
      #  print(i,j,ans)
        if(j==ind and ans[i][j]>mini+1):
            print("NO")
            flag=True
            break
        if(i==0 and j==m):
            mini=ans[i][j]
            ind=j
    if(flag):
        break
else:
    for i in range(n):
        arr[i].append(i)
    arr=sorted(arr,key=lambda x : x[1])   
    print("YES")
    for ele in arr:
        p=ans[ele[2]]
        for i in range(k):
            if(p[i]>0):
                while(p[i]>0):
                    print(i+1,end=" ")
                    p[i]-=1
        print()        