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>
using namespace std;
const long double PI = 3.1415926535897932384626433832795;
template <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  int l = v.size();
  for (int i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
void tr() { cout << endl; }
template <typename S, typename... Strings>
void tr(S x, const Strings&... rest) {
  cout << x << ' ';
  tr(rest...);
}
const int LOGN = 18;
const int N = 1 << LOGN;
const int inf = 1e9;
struct node {
  int cnt;
  void assign(int value) { cnt = value; }
  void update(int value) { cnt = value; }
  void combine(node& left, node& right) { cnt = min(left.cnt, right.cnt); }
};
int n, s, l;
int a[N], lazy[N];
node tree[2 * N];
void build(int id = 1, int l = 0, int r = n) {
  if (l + 1 == r) {
    tree[id].assign(inf);
    return;
  }
  int left = id << 1, right = left + 1, mid = (l + r) >> 1;
  build(left, l, mid);
  build(right, mid, r);
  tree[id].combine(tree[left], tree[right]);
  return;
}
void update(int index, int val, int id = 1, int l = 0, int r = n) {
  if (l + 1 == r) {
    tree[id].update(val);
    return;
  }
  int left = id << 1, right = left + 1, mid = (l + r) >> 1;
  if (index < mid)
    update(index, val, left, l, mid);
  else
    update(index, val, right, mid, r);
  tree[id].combine(tree[left], tree[right]);
}
int query(int x, int y, int id = 1, int l = 0, int r = n) {
  if (x >= r or l >= y) return inf;
  if (x <= l && r <= y) return tree[id].cnt;
  int left = id << 1, right = left + 1, mid = (l + r) >> 1;
  return min(query(x, y, left, l, mid), query(x, y, right, mid, r));
}
int sp1[LOGN][1 << LOGN], sp2[LOGN][1 << LOGN], k[1 << LOGN], o[1 << LOGN];
void sparseTable() {
  for (int i = 0; i < n; i++) {
    sp1[0][i] = sp2[0][i] = a[i];
  }
  for (int j = 1; j < LOGN; j++) {
    for (int i = 0; (i + (1 << j) - 1) < n; i++) {
      sp1[j][i] = max(sp1[j - 1][i], sp1[j - 1][i + (1 << (j - 1))]);
      sp2[j][i] = min(sp2[j - 1][i], sp2[j - 1][i + (1 << (j - 1))]);
    }
  }
  for (int i = 2; i <= n; i++) {
    k[i] = k[i - 1] + ((i & (i - 1)) == 0);
    if (k[i] == k[i - 1])
      o[i] = o[i - 1];
    else
      o[i] = ((o[i - 1] + 1) << 1) - 1;
  }
  return;
}
int mxq(int x, int y) {
  int l = y - x + 1;
  return max(sp1[k[l]][x], sp1[k[l]][y - o[l]]);
}
int mnq(int x, int y) {
  int l = y - x + 1;
  return min(sp2[k[l]][x], sp2[k[l]][y - o[l]]);
}
int get(int r) {
  int lo = -1, hi = r, mid;
  while (lo + 1 < hi) {
    mid = (lo + hi) / 2;
    if (mxq(mid, r) - mnq(mid, r) > s)
      lo = mid;
    else
      hi = mid;
  }
  return hi;
}
int main() {
  scanf("%d%d%d", &n, &s, &l);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  int mn = 1e9, mx = -1e9;
  for (int i = 0; i < l; i++) {
    mn = min(mn, a[i]);
    mx = max(mx, a[i]);
  }
  if (mx - mn > s or n < l) {
    puts("-1");
    return 0;
  }
  sparseTable();
  build();
  update(l - 1, 1);
  for (int i = l; i < n; i++) {
    int y = i - l;
    int x = get(i);
    x--;
    if (x > y) {
      update(i, inf);
    } else {
      if (x == -1) {
        update(i, 1);
        continue;
      }
      int v = query(x, y + 1);
      update(i, v + 1);
    }
  }
  int res = query(n - 1, n);
  if (res == inf) res = -1;
  printf("%d\n", res);
  return 0;
}
