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.
// Don't place your source in a package
import java.util.*;
import java.lang.*;
import java.io.*;

// Please name your class Main
public class Main {
  int n, s, l;
  int a[];
  
  class it {
    int val = 0, l, r;
    it cl, cr;
    public it (int l, int r) {
      this.l = l; this.r = r;
      if (l >= r) return ;
      int mid = (l + r) / 2;
      cl = new it(l, mid);
      cr = new it(mid + 1, r);
    }
    
    public void upd(int i, int val) {
      if (l > r || i > r || i < l) return;
      if (l == r) {
        this.val = val;
        return;
      }
      cl.upd(i, val);
      cr.upd(i, val);
      this.val = Math.min(cl.val, cr.val);
    }
    
    public int get(int ql, int qr) {
      if (l > r || l > qr || ql > r) return Integer.MAX_VALUE;
      if (ql <= l && r <= qr) return val;
      return Math.min(cl.get(ql, qr), cr.get(ql, qr));
    }
  }
  
  it ma, mi, dp;
  
  boolean ok(int l, int r) {
    if (r - l + 1 < this.l) return false;
    if (- ma.get(l, r) - mi.get(l, r) > s) return false;
    return true;
  }
  
  Main() {
    n = nxtint();
    s = nxtint();
    l = nxtint();
    a = new int[n];
    for (int i = -1; ++ i < n ; ) a[i] = nxtint();
    ma = new it(0, n - 1);
    mi = new it(0, n - 1);
    dp = new it(0, n - 1);
    for (int i = -1; ++ i < n; ) {
      ma.upd(i, -a[i]);
      mi.upd(i, a[i]);
      int l = 0, r = i - this.l + 1;
      while (l < r) {
        int mid = l + (r - l) / 2;
        if (ok(mid, i)) r = mid;
        else l = mid + 1;
      }
      if (!ok(l, i)) {
        dp.upd(i, Integer.MAX_VALUE);
      } else {
        int t = l > 0 ? dp.get(l - 1, i - this.l) : 0;
        if (t != Integer.MAX_VALUE) ++t;
        // err.println(i + " " + l + " " + t + " " + ma.get(l, i) + " " + mi.get(l, i));
        dp.upd(i, t);
      }
    }
    int ans = dp.get(n - 1, n - 1);
    if (ans == Integer.MAX_VALUE) out.println(-1);
    else out.println(ans);
  }
  
  /////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
  
  public static void main(String args[]) { new Main().out.flush(); }
  PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out), false);
  PrintStream err = System.err;
  
  BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  StringTokenizer st = null;
  String nxttok() {
    try { while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());}
    catch (IOException e) { throw new RuntimeException(e); }
    return st.nextToken();
  }
  int nxtint() { return Integer.parseInt(nxttok()); }
  long nxtlong() { return Long.parseLong(nxttok()); }
  double nxtdouble() { return Double.parseDouble(nxttok()); }

}