1256_C. Platforms Jumping
There is a river of width n. The left bank of the river is cell 0 and the right bank is cell n + 1 (more formally, the river can be represented as a sequence of n + 2 cells numbered from 0 to n + 1). There are also m wooden platforms on a river, the i-th platform has length c_i (so the i-th platform takes c_i consecutive cells of the river). It is guaranteed that the sum of lengths of platforms does not exceed n.

You are standing at 0 and want to reach n+1 somehow. If you are standing at the position x, you can jump to any position in the range [x + 1; x + d]. However you don't really like the water so you can jump only to such cells that belong to some wooden platform. For example, if d=1, you can jump only to the next position (if it belongs to the wooden platform). You can assume that cells 0 and n+1 belong to wooden platforms.

You want to know if it is possible to reach n+1 from 0 if you can move any platform to the left or to the right arbitrary number of times (possibly, zero) as long as they do not intersect each other (but two platforms can touch each other). It also means that you cannot change the relative order of platforms.

Note that you should move platforms until you start jumping (in other words, you first move the platforms and then start jumping).

For example, if n=7, m=3, d=2 and c = [1, 2, 1], then one of the ways to reach 8 from 0 is follow:

<image> The first example: n=7.

Input

The first line of the input contains three integers n, m and d (1 ≤ n, m, d ≤ 1000, m ≤ n) — the width of the river, the number of platforms and the maximum distance of your jump, correspondingly.

The second line of the input contains m integers c_1, c_2, ..., c_m (1 ≤ c_i ≤ n, ∑_{i=1}^{m} c_i ≤ n), where c_i is the length of the i-th platform.

Output

If it is impossible to reach n+1 from 0, print NO in the first line. Otherwise, print YES in the first line and the array a of length n in the second line — the sequence of river cells (excluding cell 0 and cell n + 1).

If the cell i does not belong to any platform, a_i should be 0. Otherwise, it should be equal to the index of the platform (1-indexed, platforms are numbered from 1 to m in order of input) to which the cell i belongs.

Note that all a_i equal to 1 should form a contiguous subsegment of the array a of length c_1, all a_i equal to 2 should form a contiguous subsegment of the array a of length c_2, ..., all a_i equal to m should form a contiguous subsegment of the array a of length c_m. The leftmost position of 2 in a should be greater than the rightmost position of 1, the leftmost position of 3 in a should be greater than the rightmost position of 2, ..., the leftmost position of m in a should be greater than the rightmost position of m-1.

See example outputs for better understanding.

Examples

Input


7 3 2
1 2 1


Output


YES
0 1 0 2 2 0 3 


Input


10 1 11
1


Output


YES
0 0 0 0 0 0 0 0 0 1 


Input


10 1 5
2


Output


YES
0 0 0 0 1 1 0 0 0 0 

Note

Consider the first example: the answer is [0, 1, 0, 2, 2, 0, 3]. The sequence of jumps you perform is 0 → 2 → 4 → 5 → 7 → 8.

Consider the second example: it does not matter how to place the platform because you always can jump from 0 to 11.

Consider the third example: the answer is [0, 0, 0, 0, 1, 1, 0, 0, 0, 0]. The sequence of jumps you perform is 0 → 5 → 6 → 11.
```python
from collections import defaultdict
import sys
input=sys.stdin.readline
n,m,d=map(int,input().split())
c=[int(i) for i in input().split()]
ind=defaultdict(int)
suff=n
for i in range(m-1,-1,-1):
    suff-=c[i]
    ind[i+1]=suff+1
indl=[]
for i in ind:
    indl.append(i)
indl.reverse()
cur=0
for i in indl:
    if ind[i]-cur>d:
        ind[i]=cur+d
        cur+=d
        cur+=c[i-1]-1
    else:
        cur=ind[i]+c[i-1]-1
#print(ind)
if n+1-cur>d:
    print('NO')
else:
    print('YES')
    ans=[0]*(n)
    for i in range(1,m+1):
        for j in range(ind[i]-1,min(n,ind[i]+c[i-1]-1)):
            ans[j]=i
    print(*ans)

    

```