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.File;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import static java.lang.Math.*;
import static java.util.Arrays.*;

public class Main {

	public static void main(String[] args) throws IOException {
		new Main().run();
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer st = new StringTokenizer("");

	private void run() throws IOException {
		if (new File("input.txt").exists())
			in = new BufferedReader(new FileReader("input.txt"));
		else
			in = new BufferedReader(new InputStreamReader(System.in));
		if (new File("output.txt").exists())
			out = new PrintWriter("output.txt");
		else
			out = new PrintWriter(System.out);

		solve();

		in.close();
		out.close();
	}

	final long inf = Long.MAX_VALUE / 10;
	final int MAX_N = 100 * 1000 + 100;

	int n, len, s;
	int a[] = new int[MAX_N];

	RMinQ rminq;
	RMaxQ rmaxq;
	RMQ rmqAns;

	long getMinMax(int l, int r) {
		return rmaxq.getMax(l, r) - rminq.getMin(l, r);
	}

	void solve() throws IOException {
		n = nextInt();
		s = nextInt();
		len = nextInt();
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		rminq = new RMinQ(a, n);
		rmaxq = new RMaxQ(a, n);
		rmqAns = new RMQ(n);

//		 System.err.println("this " + rmaxq.getMax(0, n - 1));
//		 System.err.println("this " + rminq.getMin(0, n - 1));
		// System.err.println(getMinMax(0, 6));

		for (int i = len - 1; i < n; i++) {
			int r = i - len + 1;
			if (getMinMax(r, i) <= s) {
				int l = -1;
				while (r - l > 1) {
					int m = (l + r) / 2;
					if (getMinMax(m, i) <= s)
						r = m;
					else
						l = m;
				}

				// System.err.println(r + " " + i);

				long minAns;
				if (r == 0)
					minAns = 0;
				else
					minAns = rmqAns.getMin(l, i - len);
				if (minAns < inf) {
					rmqAns.update(i, minAns + 1);
				}
			}
		}

		long ans = rmqAns.getMin(n - 1, n - 1);
		if (ans >= inf)
			ans = -1;
		out.println(ans);

	}

	class RMQ {
		int n;
		long treeMin[];

		public RMQ(int size) {
			n = 1;
			while (n < size)
				n <<= 1;
			treeMin = new long[n << 1];

			fill(treeMin, inf);
		}

		void update(int v, int l, int r, int index, long val) {
			if (l == r) {
				treeMin[v] = val;
				return;
			}

			int m = (l + r) >> 1;
			if (index <= m)
				update(v * 2, l, m, index, val);
			else
				update(v * 2 + 1, m + 1, r, index, val);
			treeMin[v] = min(treeMin[v * 2], treeMin[v * 2 + 1]);
		}

		void update(int index, long val) {
			update(1, 0, n - 1, index, val);
		}

		long getMin(int v, int l, int r, int ll, int rr) {
			if (rr < ll)
				return inf;

			if (l == ll && r == rr)
				return treeMin[v];

			int m = (l + r) >> 1;
			long minl = getMin(v * 2, l, m, ll, min(m, rr));
			long minr = getMin(v * 2 + 1, m + 1, r, max(m + 1, ll), rr);
			return min(minl, minr);
		}

		long getMin(int l, int r) {
			return getMin(1, 0, n - 1, l, r);
		}
	}

	class RMaxQ {
		int n;
		int logn;
		long t[][];

		public RMaxQ(int a[], int size) {
			n = size;
			logn = 0;
			while ((1 << (logn + 1)) <= n)
				logn++;

			t = new long[logn + 1][n];
			for (int i = 0; i < n; i++)
				t[0][i] = a[i];
			for (int l = 1; l <= logn; l++) {
				int l2 = 1 << l;
				for (int i = 0; i + l2 / 2 < n; i++)
					t[l][i] = max(t[l - 1][i], t[l - 1][i + l2 / 2]);
			}
		}

		long getMax(int l, int r) {
			int l2 = 0;
			int len = r - l + 1;
			while ((1 << (l2 + 1)) <= len)
				l2++;

			long maxl = t[l2][l];
			long maxr = t[l2][r - (1 << l2) + 1];
//			System.err.println(l + " " + r + " " + max(maxl, maxr));
			return max(maxl, maxr);
		}

	}

	class RMinQ {
		int n;
		int logn;
		long t[][];

		public RMinQ(int a[], int size) {
			n = size;
			logn = 0;
			while ((1 << (logn + 1)) <= n)
				logn++;

			t = new long[logn + 1][n];
			for (int i = 0; i < n; i++)
				t[0][i] = a[i];
			for (int l = 1; l <= logn; l++) {
				int l2 = 1 << l;
				for (int i = 0; i + l2 / 2 < n; i++)
					t[l][i] = min(t[l - 1][i], t[l - 1][i + l2 / 2]);
			}
		}

		long getMin(int l, int r) {
			int l2 = 0;
			int len = r - l + 1;
			while ((1 << (l2 + 1)) <= len)
				l2++;

			long minl = t[l2][l];
			long minr = t[l2][r - (1 << l2) + 1];
			return min(minl, minr);
		}

	}

	String nextLine() throws IOException {
		st = new StringTokenizer("");
		return in.readLine();
	}

	String nextToken() throws IOException {
		while (!st.hasMoreTokens()) {
			st = new StringTokenizer(in.readLine());
		}
		return st.nextToken();
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	boolean EOF() throws IOException {
		while (!st.hasMoreTokens()) {
			String str = in.readLine();
			if (str == null)
				return true;
			st = new StringTokenizer(str);
		}
		return false;
	}

}
