CF1017C The Phone Number 贪心

Mrs. Smith is trying to contact her husband, John Smith, but she forgot the secret phone number!

The only thing Mrs. Smith remembered was that any permutation of n

can be a secret phone number. Only those permutations that minimize secret value might be the phone of her husband.

The sequence of n

integers is called a permutation if it contains all integers from 1 to n

exactly once.

The secret value of a phone number is defined as the sum of the length of the longest increasing subsequence (LIS) and length of the longest decreasing subsequence (LDS).

A subsequence ai1,ai2,…,aik

where 1≤i1<i2<…<ik≤n is called increasing if ai1<ai2<ai3<…<aik. If ai1>ai2>ai3>…>aik

, a subsequence is called decreasing. An increasing/decreasing subsequence is called longest if it has maximum length among all increasing/decreasing subsequences.

For example, if there is a permutation [6,4,1,7,2,3,5]

, LIS of this permutation will be [1,2,3,5], so the length of LIS is equal to 4. LDS can be [6,4,1], [6,4,2], or [6,4,3], so the length of LDS is 3

Note, the lengths of LIS and LDS can be different.

So please help Mrs. Smith to find a permutation that gives a minimum sum of lengths of LIS and LDS.

Input

The only line contains one integer n

(1≤n≤105

) — the length of permutation that you need to build.

Output

Print a permutation that gives a minimum sum of lengths of LIS and LDS.

If there are multiple answers, print any.

Examples

Input

Copy

Output

Copy

3 4 1 2

Input

Copy

Output

Copy

2 1

Note

In the first sample, you can build a permutation [3,4,1,2]

. LIS is [3,4] (or [1,2]), so the length of LIS is equal to 2. LDS can be ony of [3,1], [4,2], [3,2], or [4,1]. The length of LDS is also equal to 2. The sum is equal to 4. Note that [3,4,1,2]

is not the only permutation that is valid.

In the second sample, you can build a permutation [2,1]

. LIS is [1] (or [2]), so the length of LIS is equal to 1. LDS is [2,1], so the length of LDS is equal to 2. The sum is equal to 3. Note that permutation [1,2] is also valid.

选取 $\sqrt{n}$ 块时，自然最小-------> 均值不等式；

对于每一块，我们可以递增排序，整体来看，递减排列即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 500005
#define inf 0x3f3f3f3f
#define INF 999999999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-7
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

int n;

int a[maxn];
int main()
{
	//ios::sync_with_stdio(false);
	rdint(n);
	int tp = sqrt(n);
	int part = n / tp;
	for (int i = 1, ed = n, j; i <= n; i += tp) {
		for (j = min(n, i + tp - 1); j >= i; j--)a[j] = ed--;
	}
	for (int i = 1; i <= n; i++)cout << a[i] << ' ';
	cout << endl;
	return 0;
}

CF1017C The Phone Number 贪心

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