Problem:
Score : 300 points

Problem Statement

Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?

Constraints

1 \leq a, b, c \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

a \ b \ c

Output

If \sqrt{a} + \sqrt{b} < \sqrt{c}, print Yes; otherwise, print No.

Sample Input 1

2 3 9

Sample Output 1

No

\sqrt{2} + \sqrt{3} < \sqrt{9} does not hold.

Sample Input 2

2 3 10

Sample Output 2

Yes

\sqrt{2} + \sqrt{3} < \sqrt{10} holds.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/*
 * WelcomeToAtCoder.cpp
 *
 *  Created on: 2020/02/19
 *      Author: black
 */
#include <iostream>
#include <math.h>
using namespace std;

typedef long double ld;
const ld eps = 1.0e-18; // 許容される誤差

ld mysqrtl(ld x)
{
  ld a = sqrt((double)x); // 近似値
  do {
    a = (a + x/a) / 2.0L;
  } while (fabsl(x - a*a) > eps);
  return a;
}

int main(){
	ld a, b, c;
	long double A, C;
	cin >> a >> b >> c;

	A = mysqrtl(a+b);
	C = mysqrtl(c);

	if (A < C){
		cout << "Yes" << endl;
	}
	else{
		cout << "No" << endl;
	}


	return 0;
}