题干：

You are going to the beach with the idea to build the greatest sand castle ever in your head! The beach is not as three-dimensional as you could have imagined, it can be decribed as a line of spots to pile up sand pillars. Spots are numbered 1 through infinity from left to right.

Obviously, there is not enough sand on the beach, so you brought n packs of sand with you. Let height hi of the sand pillar on some spot i be the number of sand packs you spent on it. You can't split a sand pack to multiple pillars, all the sand from it should go to a single one. There is a fence of height equal to the height of pillar with H sand packs to the left of the first spot and you should prevent sand from going over it.

Finally you ended up with the following conditions to building the castle:

  • h1 ≤ H: no sand from the leftmost spot should go over the fence;
  • For any  |hi - hi + 1| ≤ 1: large difference in heights of two neighboring pillars can lead sand to fall down from the higher one to the lower, you really don't want this to happen;
  • : you want to spend all the sand you brought with you.

As you have infinite spots to build, it is always possible to come up with some valid castle structure. Though you want the castle to be as compact as possible.

Your task is to calculate the minimum number of spots you can occupy so that all the aforementioned conditions hold.

Input

The only line contains two integer numbers n and H (1 ≤ n, H ≤ 1018) — the number of sand packs you have and the height of the fence, respectively.

Output

Print the minimum number of spots you can occupy so the all the castle building conditions hold.

Examples

Input

5 2

Output

3

Input

6 8

Output

3

Note

Here are the heights of some valid castles:

  • n = 5, H = 2, [2, 2, 1, 0, ...], [2, 1, 1, 1, 0, ...], [1, 0, 1, 2, 1, 0, ...]
  • n = 6, H = 8, [3, 2, 1, 0, ...], [2, 2, 1, 1, 0, ...], [0, 1, 0, 1, 2, 1, 1, 0...] (this one has 5 spots occupied)

The first list for both cases is the optimal answer, 3 spots are occupied in them.

And here are some invalid ones:

  • n = 5, H = 2, [3, 2, 0, ...], [2, 3, 0, ...], [1, 0, 2, 2, ...]
  • n = 6, H = 8, [2, 2, 2, 0, ...], [6, 0, ...], [1, 4, 1, 0...], [2, 2, 1, 0, ...]

题目大意：

     对给定的n,H，把n划分为a1,a2，a3,...a1,a2，a3,...，要求首项a1≤Ha1≤H，相邻两项之差不大于1，而且最后一项必须是1。总个数要最少，输出这个最小的总个数。

解题报告：

附一个比较好的题解链接

   总的来说，就是先把能填的填上，然后剩下的按照正态分布填（这样可以保证最大）二分验证这个可以可行即可。注意会爆longlong所以给出两种处理方式。

AC代码1：

#include<bits/stdc++.h>
#define ll long long
using namespace std;

ll n,h;
bool ok(ll x) {
	ll ans=0;
	if(x>=2e9) return 1;
	if(x<=h) ans=x*(x+1)/2;
	else {
		ans=h*(h+1)/2+(x-h)*h;	
		ll high=x-h-1;
		ll t1=high/2,t2=high-t1;
		ans+=t1*(t1+1)/2+t2*(t2+1)/2;
	}
	if(ans>=n)return 1;
	else return 0;
}
int main() 
{
	cin>>n>>h;
	ll l=1,r=1e18+1;
	ll mid=(l+r)/2;
	while(l<r) {
		mid = (l+r)/2;
		if(ok(mid)) r=mid;
		else l=mid+1;
	}
	printf("%lld\n",l);
	return 0;
}

AC代码2：

#include<bits/stdc++.h>
#define ll long long
using namespace std;

ll n,h;
bool ok(ll x) {
	ll ans=0;
//	if(x>=2e9) return 1;
	if(x<=h) ans=x*(x+1)/2;
	else {
		ans=h*(h+1)/2+(x-h)*h;	
		ll high=x-h-1;
		ll t1=high/2,t2=high-t1;
		ans+=t1*(t1+1)/2+t2*(t2+1)/2;
	}
	if(ans>=n) return 1;
	else return 0;
}
int main() 
{
	cin>>n>>h;
	ll l=1,r=2*sqrt(n)+1;
	ll mid=(l+r)/2;
	while(l<r) {
		mid = (l+r)/2;
		if(ok(mid)) r=mid;
		else l=mid+1;
	}
	printf("%lld\n",l);
	return 0;
}

总结：难啊！！难吗？？其实还是做题经验太少了。。。。