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.
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:367721060")
using namespace std;
const long long modulus = 1e9 + 7;
const long long INF = 1e18;
const long long MAX = 101000;
int n;
long long s;
long long l;
long long a[MAX];
long long dp[MAX];
struct node {
  long long keyMIN;
  long long keyMAX;
  int tl;
  int tr;
  node *left;
  node *right;
  node(long long *a, int l, int r) : tl(l), tr(r) {
    int mid = (l + r) / 2;
    if (l == r) {
      keyMIN = keyMAX = a[l];
      return;
    } else {
      left = new node(a, l, mid);
      right = new node(a, mid + 1, r);
      keyMIN = min(left->keyMIN, right->keyMIN);
      keyMAX = max(left->keyMAX, right->keyMAX);
    }
  }
};
pair<long long, long long> Query(node *v, int l, int r) {
  int mid = (v->tl + v->tr) / 2;
  if (v->tl == l && v->tr == r) {
    return make_pair(v->keyMIN, v->keyMAX);
  }
  if (mid >= r)
    return Query(v->left, l, r);
  else if (mid < l)
    return Query(v->right, l, r);
  else {
    pair<long long, long long> tmp1 = Query(v->left, l, mid);
    pair<long long, long long> tmp2 = Query(v->right, mid + 1, r);
    return make_pair(min(tmp1.first, tmp2.first),
                     max(tmp1.second, tmp2.second));
  }
}
void update(node *v, int pos, int x) {
  int mid = (v->tl + v->tr) / 2;
  if (v->tl == pos && v->tr == pos) {
    v->keyMIN = v->keyMAX = x;
    return;
  }
  if (pos <= mid)
    update(v->left, pos, x);
  else if (pos > mid)
    update(v->right, pos, x);
  v->keyMIN = min(v->left->keyMIN, v->right->keyMIN);
  v->keyMAX = max(v->left->keyMAX, v->right->keyMAX);
}
int main() {
  long long MIN = 1e9 + 7, MAX = -1e9 + 7;
  cin >> n >> s >> l;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] < MIN) MIN = a[i];
    if (a[i] > MAX) MAX = a[i];
  }
  if (n < l) {
    cout << -1 << endl;
    return 0;
  }
  if (MAX - MIN < s) {
    cout << 1 << endl;
    return 0;
  }
  node *rootMAS = new node(a, 1, n);
  node *rootDP = new node(dp, 1, n);
  for (int i = n; i >= n - l + 2; i--) {
    update(rootDP, i, 1e9 + 300);
  }
  for (int i = n - l + 1; i >= 1; i--) {
    long long good = i;
    long long bad = n + 1;
    while (bad - good > 1) {
      long long mid = (good + bad) / 2;
      pair<long long, long long> tmp = Query(rootMAS, i, mid);
      if (tmp.second - tmp.first <= s)
        good = mid;
      else
        bad = mid;
    }
    if (good - i + 1 < l) {
      update(rootDP, i, 1e9 + 300);
    } else if (good == n) {
      update(rootDP, i, 1);
    } else {
      pair<long long, long long> tmp = Query(rootDP, i + l, good + 1);
      update(rootDP, i, tmp.first + 1);
    }
  }
  pair<long long, long long> tmp = Query(rootDP, 1, 1);
  if (tmp.first > n) {
    cout << -1 << endl;
    return 0;
  } else {
    cout << tmp.first << endl;
  }
  return 0;
}
