Vasya has got an array consisting of n integers, and two integers k and len in addition. All numbers in the array are either between 1 and k (inclusive), or equal to -1. The array is good if there is no segment of len consecutive equal numbers.

Vasya will replace each -1 with some number from 1 to k (inclusive) in such a way that the resulting array is good. Tell him the number of ways to do this replacement. Since the answer may be large, print it modulo 998244353.

Input

The first line contains three integers n, k and len (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 100, 1 ≤ len ≤ n).

The second line contains n numbers — the array. Each number is either -1 or between 1 and k (inclusive).

Output

Print one integer — the number of ways to replace each -1 with some number from 1 to k (inclusive) so the array is good. The answer may be large, so print it modulo 998244353.

Examples

Input


5 2 3
1 -1 1 -1 2


Output


2


Input


6 3 2
1 1 -1 -1 -1 -1


Output


0


Input


10 42 7
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1


Output


645711643

Note

Possible answers in the first test: 

  1. [1, 2, 1, 1, 2]; 
  2. [1, 2, 1, 2, 2]. 



There is no way to make the array good in the second test, since first two elements are equal.

There are too many answers in the third test, so we won't describe any of them.
#include <bits/stdc++.h>
using namespace std;
long long n, k, len, a[100100], sum[100100], f[100100][110], e[100100][110],
    l[100100][110];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k >> len;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sum[0] = 1;
  len--;
  for (int i = 1; i <= n; i++) {
    for (int x = 1; x <= k; x++)
      if (x == a[i] || a[i] == -1) {
        f[i][x] = sum[i - 1];
        f[i][x] = (f[i][x] - e[i - 1][x] + 998244353) % 998244353;
        l[i][x] = l[i - 1][x] + 1;
        if (l[i][x] > len) l[i][x] = len;
        if (l[i][x] == len)
          e[i][x] = (sum[i - len] - f[i - len][x] + 998244353) % 998244353;
        sum[i] = (sum[i] + f[i][x]) % 998244353;
      }
  }
  if (len != 0)
    cout << sum[n] << endl;
  else
    cout << 0 << endl;
  return 0;
}
