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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class codeforce {
	public static PrintWriter out;
    
	   //-----------MyScanner class for faster input----------
	   public static class MyScanner {
	      BufferedReader br;
	      StringTokenizer st;
	 
	      public MyScanner() {
	         br = new BufferedReader(new InputStreamReader(System.in));
	      }
	 
	      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());
	      }
	 
	      double nextDouble() {
	          return Double.parseDouble(next());
	      }
	 
	      String nextLine(){
	          String str = "";
		  try {
		     str = br.readLine();
		  } catch (IOException e) {
		     e.printStackTrace();
		  }
		  return str;
	      }

	   }
	   public static void main(String args[]) throws Exception {
			class prob {
				void GenSparseTable(int[][] mx, int[][] mn, int[] in) {
					int n = in.length;
					int sq = (int) (Math.log(n)/Math.log(2));
					
					for (int j = 1; j <= sq; ++j) {
						for (int i = 0; i + (1 << j) - 1 < n; i++) {
				              if (mx[i][j - 1] > mx[i + (1 << (j - 1))][j - 1])
				                  mx[i][j] = mx[i][j - 1];
				              else
				                  mx[i][j] = mx[i + (1 << (j - 1))][j - 1];
				              if (mn[i][j - 1] < mn[i + (1 << (j - 1))][j - 1])
				                  mn[i][j] = mn[i][j - 1];
				              else
				                  mn[i][j] = mn[i + (1 << (j - 1))][j - 1];
						}
					}
				}
				int query(int[][] mx, int[][] mn, int l , int r) {
					// 2^x >= (r - l) + 1
					int k = (int) (Math.log(r - l + 1)/ Math.log(2));
					return Math.max(mx[l][k], mx[r - (1 << k) + 1][k]) - Math.min(mn[l][k], mn[r - (1 << k) + 1][k]);
				}
				public void solve() {
					MyScanner sc = new MyScanner();
					int n = sc.nextInt();
					int s = sc.nextInt();
					int l = sc.nextInt();
					int sq = (int) (Math.log(n)/Math.log(2));
					int[] in = new int[n];
					int[][] mx = new int[n][sq + 1];
					int[][] mn = new int[n][sq + 1];
					
					for (int i = 0; i < n; ++i) {
						in[i] = sc.nextInt();
						mx[i][0] = in[i];
						mn[i][0] = in[i];
					}
					GenSparseTable(mx, mn, in);
					int out[] = new int[n + 1];
					
					Arrays.fill(out, Integer.MAX_VALUE);
					out[0] = 0;
					int count = 0;

					for (int i = l; i <= n; ++i) {
						while (i - count >= l && (query(mx, mn, count, i - 1) > s || out[count] == Integer.MAX_VALUE)) {
							++count;
						}
						if (i - count >= l) {
							out[i] = Math.min(out[i], out[count] + 1);
						}
					}
					
					System.out.println(out[n] == Integer.MAX_VALUE ? -1 : out[n]);
				}
			}
			prob s = new prob();
			s.solve();
		}
	}
