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;
struct edge {
  long long mx, mn;
};
long long arr[100005];
int dptree[400005];
struct edge tree[4 * 100005];
int l, n, limit;
void build(int node, int a, int b) {
  if (a == b) {
    tree[node].mx = arr[a];
    tree[node].mn = arr[a];
    return;
  }
  int mid = (a + b) >> 1;
  build((node << 1), a, mid);
  build((node << 1) | 1, mid + 1, b);
  tree[node].mx = max(tree[(node << 1)].mx, tree[(node << 1) | 1].mx);
  tree[node].mn = min(tree[(node << 1)].mn, tree[(node << 1) | 1].mn);
}
struct edge query(int node, int a, int b, int i, int j) {
  if (a >= i && b <= j) return tree[node];
  int mid = (a + b) >> 1;
  if (j <= mid)
    return query((node << 1), a, mid, i, j);
  else if (i > mid)
    return query((node << 1) | 1, mid + 1, b, i, j);
  else {
    struct edge p1, p2, p;
    p1 = query((node << 1), a, mid, i, j);
    p2 = query((node << 1) | 1, mid + 1, b, i, j);
    p.mx = max(p1.mx, p2.mx);
    p.mn = min(p1.mn, p2.mn);
    return p;
  }
}
int dpquery(int node, int a, int b, int i, int j) {
  if (a >= i && b <= j) return dptree[node];
  int mid = (a + b) >> 1;
  if (j <= mid)
    return dpquery((node << 1), a, mid, i, j);
  else if (i > mid)
    return dpquery((node << 1) | 1, mid + 1, b, i, j);
  else
    return min(dpquery((node << 1), a, mid, i, j),
               dpquery((node << 1) | 1, mid + 1, b, i, j));
}
void update(int node, int a, int b, int i, int v) {
  if (a == b) {
    dptree[node] = v;
    return;
  }
  int mid = (a + b) >> 1;
  if (i <= mid)
    update((node << 1), a, mid, i, v);
  else
    update((node << 1) | 1, mid + 1, b, i, v);
  dptree[node] = min(dptree[(node << 1)], dptree[(node << 1) | 1]);
}
int bin_search(int v, int i, int j) {
  if (i >= j) return j;
  int mid = ceil((i + j) / 2.0);
  struct edge tmp = query(1, 0, n - 1, v, mid);
  if (tmp.mx - tmp.mn <= limit)
    return bin_search(v, mid, j);
  else
    return bin_search(v, i, mid - 1);
}
int till[100005];
int dp[100005];
int main() {
  int i;
  scanf("%d", &n);
  scanf("%d", &limit);
  scanf("%d", &l);
  for (i = 0; i < n; i++) {
    scanf("%I64d", &arr[i]);
  }
  build(1, 0, n - 1);
  for (i = 0; i < n; i++) {
    till[i] = bin_search(i, i, n - 1);
  }
  fill(dptree, dptree + 4 * n, 1000000007);
  update(1, 0, n, n, 0);
  for (i = n - 1; i >= 0; i--) {
    int fr = i + l;
    int to = till[i] + 1;
    if (fr > to)
      dp[i] = 1000000007;
    else {
      dp[i] = 1 + dpquery(1, 0, n, fr, to);
      update(1, 0, n, i, dp[i]);
    }
  }
  if (dp[0] < 1000000007)
    printf("%d\n", dp[0]);
  else
    printf("-1\n");
  return 0;
}
