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.*;
import java.util.*;
import java.math.BigInteger;

public class Main{
    final boolean isFileIO =  false;
    BufferedReader in;
    PrintWriter out;
    StringTokenizer st = new StringTokenizer("");
    String delim = " ";

    public static void main(String[] args) throws IOException {
        Main m = new Main();
        m.initIO();
        m.solve();
        m.in.close();
        m.out.close();
    }

    public void initIO() throws IOException {
        if(!isFileIO) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        } else {
            in = new BufferedReader(new FileReader("strings.in"));
            out = new PrintWriter("strings.out"); 
        }
    }
  
    String nextToken() throws IOException {
        if(!st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken(delim);
    }
 
    String readLine() throws IOException {
        return in.readLine();
    }

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

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

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

    public interface Operation {
        public int op(int a, int b);
    }

    public class SparseTable {
        int[][] b;
        int[] logs;
        int n;
        int m;
        Operation op;
        public SparseTable(int[] a, Operation op) {
            n = a.length;
            this.op = op;
            logs = new int[n + 1];
            for(int i = 0; i <= n; i++) {
                int log = 0;
                for(int j = 0; j <= 20; j++) {
                    if((1 << j) <= i) {
                        log = j;
                    }
                }
                logs[i] = log;
            }
            m = logs[n];
            b = new int[m + 1][n];
            for(int i = 0; i < n; i++) {
                b[0][i] = a[i];
            }
            for(int i = 1; i <= m; i++) {
                int r = (1 << i);
                for(int j = 0; j + r <= n; j++) {
                    b[i][j] = op.op(b[i - 1][j], b[i - 1][j + r / 2]);
                }
            }
        }

        public int rmq(int l, int r) {
            int diff = (r - l + 1);
            int p = logs[diff];
            return this.op.op(b[p][l], b[p][r - (1 << p) + 1]);
        }

    }

    public class SegmentTree {
        int[] b;
        int size;
        public SegmentTree(int[] a) {
            int n = a.length;
            int log = 0;
            for(int i = 0; i <= 20; i++) {
                if((1 << i) <= n) {
                    log = i;
                }
            }
            size = (1 << (log + 1));
            int all = size * 2 - 1;
            b = new int[all];
            for(int i = size - 1; i < size + n - 1; i++) {
                b[i] = a[i - size + 1];
            }
            for(int i = size + n - 1; i < all; i++) {
                b[i] = Integer.MAX_VALUE;
            }
            for(int i = size - 2; i >= 0; i--) {
                b[i] = Math.min(b[i * 2 + 1], b[i * 2 + 2]);
            }
        }

        public int rmq(int left, int right, int q_l, int q_r, int node) {
            if(q_l > q_r) return Integer.MAX_VALUE;
            if(left == q_l && right == q_r) {
                return b[node];
            }
            int mid = left + (right - left) / 2;
            return Math.min(rmq(left, mid, q_l, Math.min(mid, q_r), 2 * node + 1),
                rmq(mid + 1, right, Math.max(mid + 1, q_l), q_r, 2 * node + 2));
        }

        public void update(int left, int right, int pos, int v, int node) {
            if(left == right) {
                b[node] = v;
                return;
            }
            int mid = left + (right - left) / 2;
            if(pos <= mid) {
                update(left, mid, pos, v, node * 2 + 1);
            } else {
                update(mid + 1, right, pos, v, node * 2 + 2);
            }
            b[node] = Math.min(b[node * 2 + 1], b[node * 2 + 2]);
        }

    }

    public void solve() throws IOException {
        int n, s, l;
        n = nextInt(); s = nextInt(); l = nextInt();
        int[] a = new int[n];
        for(int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        SparseTable st_min = new SparseTable(a, new Operation() {
            public int op(int a, int b) {
                return Math.min(a, b);
            }
        });
        SparseTable st_max = new SparseTable(a, new Operation() {
            public int op(int a, int b) {
                return Math.max(a, b);
            }
        });
        int[] f = new int[n];
        Arrays.fill(f, Integer.MAX_VALUE);
        SegmentTree seg = new SegmentTree(f);
        for(int i = l - 1; i < n; i++) {
            int left, mid, right, min_val, max_val;
            right = i - l + 1;
            left = min_val = max_val = mid = 0;
            while(left < right) {
                mid = left + (right - left) / 2;
                min_val = st_min.rmq(mid, i);
                max_val = st_max.rmq(mid, i);
                if(max_val - min_val > s) {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            }
            max_val = st_max.rmq(left, i);
            min_val = st_min.rmq(left, i);
            if(max_val - min_val <= s) {
                int min = seg.rmq(0, seg.size - 1, Math.max(left - 1, 0), Math.max(i - l, 0), 0);
                if(left - 1 < 0) {
                    f[i] = 1;
                } else {
                    if(min != Integer.MAX_VALUE) {
                        f[i] = min + 1;    
                    }
                }
            }
            seg.update(0, seg.size - 1, i, f[i], 0);
        }
        if(f[n - 1] == Integer.MAX_VALUE) {
            out.println(-1);
        } else {
            out.println(f[n - 1]);
        }
    }

}

class Utils {

    public static long binpow(long a, long exp, long mod) {
        if(exp == 0) {
            return 1;
        }
        if(exp % 2 == 0) {
            long temp = binpow(a, exp / 2, mod);
            return (temp * temp) % mod;
        } else {
            return (binpow(a, exp - 1, mod) * a) % mod;
        }
    }

    public static long inv(long a, long mod) {
        return binpow(a, mod - 2, mod);
    }

    public static long addmod(long a, long b, long mod) {
        return ((a + b) % mod + mod) % mod;
    }

    public static long gcd(long a, long b) {
        if(b == 0)
            return a;
        return gcd(b, a % b);
    }

    //mul must be < 10^18
    public static long mulmod(long a, long b, long mod) {
        return (a * b + (((a * b) / mod) + 1) * mod) % mod;
    }
}
