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.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        new Main().run(in, out);
        out.close();
    }

    public static long mod = 17352642619633L;

    int N;
    int L;
    long S;
    long[] a;
    void run(FastScanner in, PrintWriter out) {

        N = in.nextInt();
        S = in.nextLong();
        L = in.nextInt();

        a = new long[N];
        for (int i = 0; i < N; i++) {
            a[i] = in.nextLong();
        }

        // g[i] = length of substrip such that i is the right bound
        // the length of substrip g[i+1] can never go past the left bound
        // of g[i], otherwise g[i] could also go past that bound
        // so we can use a monotonic queue to keep min/max
        int[] g = new int[N];
        MinMaxQueue mmq = new MinMaxQueue();
        for (int l = 0, r = 0; r < N; r++) {
            mmq.offerLast(a[r]);
            while (l < r && mmq.getMax()-mmq.getMin() > S) {
                mmq.pollFirst(a[l++]);
            }
            g[r] = r-l+1;
        }

        // dp[i] = min number of substrips from 0...i
        mmq.clear();
        int[] dp = new int[N+1];

        Arrays.fill(dp, -1);
        dp[0] = 0;

        int ql, qr;

        qr = 0;
        ql = -1;
        for (int i = L-1; i < N; i++) {
            qr = i-L;
            if (qr >= ql && dp[qr+1] != -1) mmq.offerLast(dp[qr+1]);

            while (ql < i-g[i] && !mmq.minQueue.isEmpty()) {
                mmq.pollFirst(dp[ql+1]);
                ql++;
            }

            // System.out.println(i + " " + ql + " " + qr + " " + Arrays.toString(mmq.minQueue.toArray()));

            if (!mmq.minQueue.isEmpty()) {
                dp[i+1] = (int) mmq.getMin() + 1;
            }
        }

        // System.out.println(Arrays.toString(dp));
        out.println(dp[N] == 0 ? -1 : dp[N]);
    }

    class MinMaxQueue {
        LinkedList<Long> maxQueue = new LinkedList<>();
        LinkedList<Long> minQueue = new LinkedList<>();
        void clear() {
            minQueue.clear();
            maxQueue.clear();
        }

        void offerLast(long x) {
            while (!maxQueue.isEmpty() && x > maxQueue.peekLast()) {
                maxQueue.pollLast();
            }
            maxQueue.offerLast(x);

            while (!minQueue.isEmpty() && x < minQueue.peekLast()) {
                minQueue.pollLast();
            }
            minQueue.offerLast(x);
        }

        void pollFirst(long x) {

            if (maxQueue.peekFirst() == x) maxQueue.pollFirst();
            if (minQueue.peekFirst() == x) minQueue.pollFirst();
        }
        long getMax() {
            return maxQueue.peekFirst();
        }
        long getMin() {
            return minQueue.peekFirst();
        }
    }


    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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