Alexandra has a paper strip with n numbers on it. Let's call them ai from left to right.

Now Alexandra wants to split it into some pieces (possibly 1). For each piece of strip, it must satisfy:

  * Each piece should contain at least l numbers.
  * The difference between the maximal and the minimal number on the piece should be at most s.



Please help Alexandra to find the minimal number of pieces meeting the condition above.

Input

The first line contains three space-separated integers n, s, l (1 ≤ n ≤ 105, 0 ≤ s ≤ 109, 1 ≤ l ≤ 105).

The second line contains n integers ai separated by spaces ( - 109 ≤ ai ≤ 109).

Output

Output the minimal number of strip pieces.

If there are no ways to split the strip, output -1.

Examples

Input

7 2 2
1 3 1 2 4 1 2


Output

3


Input

7 2 2
1 100 1 100 1 100 1


Output

-1

Note

For the first sample, we can split the strip into 3 pieces: [1, 3, 1], [2, 4], [1, 2].

For the second sample, we can't let 1 and 100 be on the same piece, so no solution exists.
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) {
		new Task().solve();
	}
}

class Task {
	InputReader in = new InputReader(System.in);
	PrintWriter out = new PrintWriter(System.out);
	final int MAXN = 100010;
	int n ;
	int[][] cmax = new int[MAXN][20];
	int[][] cmin = new int[MAXN][20];
	int maxx, minx;
	int[] a = new int[MAXN];
	int[] tlog = new int[MAXN];
	{
		tlog[0] = tlog[1] = 0;
		for (int i = 2; i < MAXN; i++)
			tlog[i] = tlog[i >> 1] + 1;
	}

	void RMQ() {
		for (int i = 1; i <= n; i++)
			cmax[i][0] = cmin[i][0] = a[i];

		for (int j = 1; (1 << j) <= n; j++) {
			for (int i = 1; i + (1 << j) - 1 <= n; i++) {
				cmax[i][j] = Math.max(cmax[i][j - 1] , cmax[i + (1 << (j - 1))][j - 1]);
				cmin[i][j] = Math.min(cmin[i][j - 1] , cmin[i + (1 << (j - 1))][j - 1]);
			}
		}
	}

	int ask(int l, int r) {
		int i = tlog[r - l + 1];
		return Math.max(cmax[l][i], cmax[r - (1 << i) + 1][i]) - Math.min(cmin[l][i], cmin[r - (1 << i) + 1][i]) ;
	}

	int[] dp = new int[MAXN] ;
	
	void solve() {
		n = in.nextInt() ;
		int s = in.nextInt() ;
		int l = in.nextInt() ; 
		for(int i = 1 ; i <= n ; i++) a[i] = in.nextInt() ; 
		RMQ() ; 
		Arrays.fill(dp , 0 , n+1 , Integer.MAX_VALUE) ;
		int partition = 0 ; 
		dp[0] = 0 ;
		for(int i = l ; i <= n ; i++){
			while(dp[partition] == Integer.MAX_VALUE ||
					(i - partition >= l && ask(partition+1 , i) > s)) 
				partition++ ;
			if(i - partition >= l && dp[partition] != Integer.MAX_VALUE)
				dp[i] = dp[partition] + 1 ;
		}
		out.println(dp[n] == Integer.MAX_VALUE ? -1 : dp[n]) ;
		out.flush() ; 
	}

}

class InputReader {
	public BufferedReader reader;
	public StringTokenizer tokenizer;

	public InputReader(InputStream stream) {
		reader = new BufferedReader(new InputStreamReader(stream), 32768);
		tokenizer = new StringTokenizer("");
	}

	private void eat(String s) {
		tokenizer = new StringTokenizer(s);
	}

	public String nextLine() {
		try {
			return reader.readLine();
		} catch (Exception e) {
			return null;
		}
	}

	public boolean hasNext() {
		while (!tokenizer.hasMoreTokens()) {
			String s = nextLine();
			if (s == null)
				return false;
			eat(s);
		}
		return true;
	}

	public String next() {
		hasNext();
		return tokenizer.nextToken();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public BigInteger nextBigInteger() {
		return new BigInteger(next());
	}

}
