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.Scanner;
import java.util.Deque;
import java.util.ArrayDeque;


public class B {
	final int oo = (int)1e9;
	class IT {
		int[] low, high;
		int[] min, max;
		IT(int N) {
			low = new int[4*N+4];
			high = new int[4*N+4];
			min = new int[4*N+4];
			max = new int[4*N+4];
			init(1, 0, N-1);
		}
		private void init(int i, int l, int h) {
			low[i] = l;
			high[i] = h;
			min[i] = oo;
			max[i] = -oo;
			if (l<h) {
				int mid = (h-l)/2+l;
				init(i*2, l, mid);
				init(i*2+1, mid+1, h);
			}
		}
		int getMin(int l, int h) {
			return getMin(1, l, h);
		}
		private int getMax(int i, int l, int h) {
			if (l > high[i] || h < low[i])
				return -oo;
			if (l <= low[i] && h >= high[i])
				return max[i];
			return Math.max(getMax(i*2, l, h), getMax(i*2+1, l, h));
		}
		int getMax(int l, int h) {
			return getMax(1, l, h);
		}
		private int getMin(int i, int l, int h) {
			if (l > high[i] || h < low[i])
				return oo;
			if (l <= low[i] && h >= high[i])
				return min[i];
			return Math.min(getMin(i*2, l, h), getMin(i*2+1, l, h));
		}
		private void set(int i, int p, int val) {
			if (low[i] > p || high[i] < p)
				return;
			if (low[i] == high[i]) {
				min[i] = val;
				max[i] = val;
			} else {
				set(i*2, p, val);
				set(i*2+1, p, val);
				min[i] = Math.min(min[i*2], min[i*2+1]);
				max[i] = Math.max(max[i*2], max[i*2+1]);
			}
		}
		void set(int p, int val) {
			set(1, p, val);
		}
	}
	class MinMaxQueue {
		Deque<Integer> q = new ArrayDeque<Integer>();
		Deque<Integer> min = new ArrayDeque<Integer>();
		Deque<Integer> max = new ArrayDeque<Integer>();
		void offer(int val) {
			q.offer(val);
			while(!min.isEmpty() && min.peekLast() > val)
				min.pollLast();
			min.offer(val);
			while(!max.isEmpty() && max.peekLast() < val)
				max.pollLast();
			max.offer(val);
		}
		int poll() {
			int ret = q.poll();
			if (min.peek() == ret)
				min.poll();
			if (max.peek() == ret)
				max.poll();
			return ret;
		}
		int getMin() {
			if (min.isEmpty())
				return oo;
			return min.peek();
		}
		int getMax() {
			if (max.isEmpty())
				return -oo;
			return max.peek();
		}
	}
	B(Scanner in) {
		int N = in.nextInt();
		int S = in.nextInt();
		int L = in.nextInt();
		IT valsIT = new IT(N);
		int[] vals = new int[N];
		for (int i=0; i<N; i++) {
			vals[i] = in.nextInt();
			valsIT.set(i, vals[i]);
		}
		IT ans = new IT(N+1);
		ans.set(0, 0);
		MinMaxQueue q = new MinMaxQueue();
		int min = 0;
		for (int i=0; i<N; i++) {
			q.offer(vals[i]);
			while(q.getMax() - q.getMin() > S) {
				min++;
				q.poll();
			}
			int curAns = ans.getMin(min, i-L+1)+1;
			ans.set(i+1, curAns);
//			System.out.println(max + " " + i + " " + curAns);
		}
		int fans = ans.getMin(N, N);
		if (fans >= oo)
			fans = -1;
		System.out.println(fans);
	}
	public static void main(String[] args) {
		new B(new Scanner(System.in));
	}
}
