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;
int s, n;
int l;
const int MAX_N = 120000;
int rmq1n[MAX_N * 4];
int rmq1m[MAX_N * 4];
int rmq[MAX_N * 4];
int go[MAX_N];
int dp[MAX_N];
int arr[MAX_N];
void build(int v, int tl, int tr) {
  if (tl + 1 == tr) {
    rmq1n[v] = arr[tl];
    rmq1m[v] = arr[tl];
    return;
  }
  int m = (tl + tr) >> 1;
  build(v * 2 + 1, tl, m);
  build(v * 2 + 2, m, tr);
  rmq1n[v] = min(rmq1n[v * 2 + 1], rmq1n[v * 2 + 2]);
  rmq1m[v] = max(rmq1m[v * 2 + 1], rmq1m[v * 2 + 2]);
}
int getm(int v, int tl, int tr, int l, int r) {
  if (r <= tl || tr <= l) return -1000000000;
  if (l <= tl && tr <= r) return rmq1m[v];
  int m = (tl + tr) >> 1;
  return max(getm(v * 2 + 1, tl, m, l, r), getm(v * 2 + 2, m, tr, l, r));
}
int getn(int v, int tl, int tr, int l, int r) {
  if (r <= tl || tr <= l) return 1000000000;
  if (l <= tl && tr <= r) return rmq1n[v];
  int m = (tl + tr) >> 1;
  return min(getn(v * 2 + 1, tl, m, l, r), getn(v * 2 + 2, m, tr, l, r));
}
bool ch(int l, int r) {
  if (getm(0, 0, n, l, r) - getn(0, 0, n, l, r) > s)
    return false;
  else
    return true;
}
void upd(int v, int tl, int tr, int x) {
  if (tl + 1 == tr) {
    rmq[v] = dp[x];
    return;
  }
  int m = (tl + tr) >> 1;
  if (x < m)
    upd(v * 2 + 1, tl, m, x);
  else
    upd(v * 2 + 2, m, tr, x);
  rmq[v] = min(rmq[v * 2 + 1], rmq[v * 2 + 2]);
}
void upd(int x) { upd(0, 0, n, x); }
int get(int v, int tl, int tr, int l, int r) {
  if (r <= tl || tr <= l) return 1000000000;
  if (l <= tl && tr <= r) return rmq[v];
  int m = (tl + tr) >> 1;
  return min(get(v * 2 + 1, tl, m, l, r), get(v * 2 + 2, m, tr, l, r));
}
int get(int l, int r) { return get(0, 0, n, l, r); }
int main() {
  scanf("%d%d%d", &n, &s, &l);
  for (int i = 0; i < n; ++i) scanf("%d", &arr[i]);
  build(0, 0, n);
  go[0] = 0;
  int lll = 0;
  for (int i = 1; i < n; ++i) {
    while (!ch(lll, i + 1)) ++lll;
    go[i] = lll;
  }
  dp[0] = 0;
  for (int i = 0; i < n; ++i) {
    if (i - go[i] + 1 < l) {
      dp[i + 1] = 1000000000;
      upd(i + 1);
      continue;
    }
    dp[i + 1] = get(go[i], i - l + 2) + 1;
    upd(i + 1);
  }
  if (dp[n] > n)
    cout << -1;
  else
    cout << dp[n];
  return 0;
}
