Description

Mishka received a gift of multicolored pencils for his birthday! Unfortunately he lives in a monochrome world, where everything is of the same color and only saturation differs. This pack can be represented as a sequence \(a_1, a_2, ..., a_n\) of \(n\) integer numbers — saturation of the color of each pencil.

Now Mishka wants to put all the mess in the pack in order. He has an infinite number of empty boxes to do this. He would like to fill some boxes in such a way that:

• Each pencil belongs to exactly one box;
• Each non-empty box has at least \(k\) pencils in it;
• If pencils \(i\) and \(j\) belong to the same box, then \(|a_i - a_j| \le d\), where \(|x|\) means absolute value of \(x\). Note that the opposite is optional, there can be pencils \(i\) and \(j\) such that \(|a_i - a_j| \le d\) and they belong to different boxes.

Help Mishka to determine if it's possible to distribute all the pencils into boxes. Print "YES" if there exists such a distribution. Otherwise print "NO".

Input

The first line contains three integer numbers \(n\), \(k\) and \(d\) (\(1 \le k \le n \le  5\cdot10^5, 0 \le d \le 10^9\)) — the number of pencils, minimal size of any non-empty box and maximal difference in saturation between any pair of pencils in the same box, respectively.

The second line contains \(n\) integer numbers \(a_1, a_2, \dots, a_n (1 \le a_i \le 10^9)\) — saturation of color of each pencil.

Output

Print "YES" if it's possible to distribute all the pencils into boxes and satisfy all the conditions. Otherwise print "NO".

Examples

input

6 3 10
7 2 7 7 4 2

output

YES

input

6 2 3
4 5 3 13 4 10

output

YES

input

3 2 5
10 16 22

output

NO

Note

In the first example it is possible to distribute pencils into 2 boxes with 3 pencils in each with any distribution. And you also can put all the pencils into the same box, difference of any pair in it won't exceed 10.

In the second example you can split pencils of saturations [4, 5, 3, 4] into 2 boxes of size 2 and put the remaining ones into another box.

Solution

题意：给\(n\)个数，同时给定\(k\)和\(d\)，问能否将这\(n\)个数分成若干个集合，每个集合大小至少为\(k\)，且同一个集合中的任何两个元素之差不超过\(d\)

同一个集合中的任何两个元素之差不超过\(d\)这一条件相当于集合中最大最小元素之差不超过\(d\)，所以很自然地想到要先对数组\(a\)排序，然后用\(f[i]\)表示前\(i\)个元素能否合法地划分，\(a[i]\)所在的集合对应着\(a_{1}, a_{2}, \dots, a_{i}\)的一个后缀，可以通过二分找到这一后缀的所有可能开始的位置，只要有一个可能的开始位置\(p\)满足\(f[p-1] = 1\)，则\(f[i] = 1\)，否则\(f[i] = 0\)，复杂度\(O(n \log n)\)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 500011;
int a[maxn], f[maxn], sum[maxn];
int main() {int n, k, d;scanf("%d%d%d", &n, &k, &d);for (int i = 1; i <= n; ++i)  scanf("%d", a + i);sort(a + 1, a + n + 1);sum[0] = 1;for (int i = 1; i <= n; ++i) {int p = lower_bound(a + 1, a + 1 + n, a[i] - d) - a;int p1 = p - 1, p2 = i - k; // [p1, p2]是可能的开始位置if (p2 >= p1) {f[i] = sum[p2] - (p1 > 0 ? sum[p1 - 1] : 0) > 0;}sum[i] = sum[i - 1] + f[i];}puts(f[n] ? "YES" : "NO");return 0;
}