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
n,k1=list(map(int,input().rstrip().split()))
a=list(map(int,input().rstrip().split()))
dp=[[-1]*100 for i in range(100)]
for i in range(n):
    for j in range(a[i]):
        dp[i][j]=0
for i in range(1,2):

    index=-1
    mini=float('inf')
    for j in range(n):
        count=0
        for k in range(100):
            if dp[j][k]==0:
                count+=1
            elif dp[j][k]==-1:
                break
        if count<mini and count!=0:
            index=j
            mini=count
    count=0
    if index==-1:
        continue
    #print(index,mini)
    for k in range(100):
        if dp[index][k]==0 and count<mini:
            dp[index][k]=i
            count+=1
        elif count==mini:
            break
    
    for j in range(n):
        count=0
        for k in range(100):
            if dp[j][k]==0 and count<=mini:
                dp[j][k]=i
                count+=1
            elif count==mini:
                break
y=max(a)
y1=min(a)
r=1
for i in range(y1+1,y):
    if r<k1:
        r+=1
    else:
        break
    for j in range(n):
        dp[j][i]=r
        
        
flag=0
for i in range(n):
    for j in range(a[i]):
        if dp[i][j]==0:
            flag=1
            break
    if flag==1:
        break
if flag==1:
    print('NO')
else:
    print('YES')
    for i in range(n):
        for j in range(a[i]):
            print(dp[i][j],end=' ')
        print()