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.util.*;
import java.math.*;
import java.io.*;

public class CF487B {
	class RMQ {
		int n, lg, sign;
		int[][] table;
		public RMQ(int nn, int ss) { // -1 for min, 1 for max
			sign = ss;
			lg = 32 - Integer.numberOfLeadingZeros(n = nn);
			table = new int[lg][n];
		}
		void init(int[] a) {
			for(int i = 0 ; i < n ; i++) table[0][i] = sign * a[i];
			for(int k = 1, len = 1 ; k < lg ; k++, len <<= 1)
				for(int i = 0 ; i < n ; i++)
					table[k][i] = Math.max(table[k - 1][i], table[k - 1][Math.min(i + len, n - 1)]);
		}
		void init(int i, int x) {
			table[0][i] = sign * x;
			for(int k = 1, len = 1 ; k < lg ; k++, len <<= 1)
				table[k][i] = Math.max(table[k - 1][i], table[k - 1][Math.min(i + len, n - 1)]);
		}
		int query(int i, int j) {
			int exp = Integer.highestOneBit(j - i + 1);
			int k = Integer.numberOfTrailingZeros(exp);
			return sign * Math.max(table[k][i], table[k][j - exp + 1]);
		}
	}
	final int oo = Integer.MAX_VALUE;
	int n, d, L;
	RMQ max, min;
	boolean good(int l, int r) { return max.query(l, r) - min.query(l, r) <= d; }
	public CF487B() {
		FS scan = new FS();
		n = scan.nextInt();  d = scan.nextInt();  L = scan.nextInt();
		int[] a = new int[n];
		for(int i = 0 ; i < n ; i++) a[i] = scan.nextInt();
		max = new RMQ(n, 1);
		min = new RMQ(n, -1);
		max.init(a);
		min.init(a);
		RMQ dpmin = new RMQ(n, -1);
		for(int i = n - 1 ; i > Math.max(-1, n - L - 1) ; i--) {
			if(n - i >= L && good(i, n - 1)) dpmin.init(i, 1);
			else dpmin.init(i, oo);
		}
		for(int i = n - L - 1 ; i >= 0 ; i--) {
			if(!good(i, i + L - 1)) {
				dpmin.init(i, oo);
				continue;
			}
			int lo = i + L - 1, hi = n - 1;
			while(lo < hi - 1) {
				int mid = (lo + hi >> 1);
				if(good(i, mid)) lo = mid;
				else hi = mid;
			}
			if(good(i, hi)) lo = hi;
			if(lo + 1 < n && dpmin.query(i + L, Math.min(lo + 1, n - 1)) == oo) {
				dpmin.init(i, oo);
				continue;
			}
			if(lo + 1 == n) dpmin.init(i, 1);
			else dpmin.init(i, dpmin.query(i + L, Math.min(lo + 1, n - 1)) + 1);
		}
		int res = dpmin.query(0, 0);
		if(res == oo) System.out.println(-1);
		else System.out.println(res);
	}
	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		public String next() {
			while (!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch (Exception e) { e.printStackTrace(); }
			}
			return st.nextToken();
		}
		public int nextInt() { return Integer.parseInt(next()); }
	}
	public static void main(String[] args) { new CF487B(); }
}
