The ultimate battle has begun in Tekken.  
Each fight has two players. Let the initial health of the players be X and Y, then, after the fight, the health of both the players reduces by \texttt{min(X, Y)}.

Anna, Bob, and Claudio have initial health levels of A, B, and C respectively. Each pair of players fight exactly once.  
Considering that you can schedule the fight between each pair in any order, find whether Anna can have a positive health level at the end of the battle.

------ Input Format ------ 

- The first line of input will contain a single integer T, denoting the number of test cases.
- The first and only line of each test case contains three space-separated integers A, B,  and C — the initial health levels of Anna, Bob, and Claudio respectively.

------ Output Format ------ 

For each test case, output on a new line, YES, if you can schedule the fight between each pair in an order such that Anna can have a positive health at the end of the battle. Otherwise, output NO.

You can print each character in uppercase or lowercase. For example NO, no, No, and nO are all considered identical.

------ Constraints ------ 

$1 ≤ T ≤ 1000$
$0 ≤ A, B, C ≤ 1000$

----- Sample Input 1 ------ 
3
4 2 0
2 7 4
5 1 4

----- Sample Output 1 ------ 
YES
NO
YES

----- explanation 1 ------ 
Test case $1$: Consider the following sequence of fights:
- Fight $1$: Anna and Bob fight. Thus, the health level of both players reduces by $\texttt{min(4, 2)} = 2$. The final health levels of Anna and Bob after the fight are $2$ and $0$ respectively.
- Fight $2$: Bob and Claudio fight. Thus, the health level of both players reduces by $\texttt{min(0, 0)} = 0$. The final health levels of Bob and Claudio after the fight are $0$ and $0$ respectively.
- Fight $3$: Anna and Claudio fight. Thus, the health level of both players reduces by $\texttt{min(2, 0)} = 0$. The final health levels of Anna and Claudio after the fight are $2$ and $0$ respectively.

Thus, in the above sequence of fights, Anna has a positive health remaining at the end.

Test case $2$: It can be shown that there exists no sequence of fights such that Anna has a positive health remaining at the end of the battle.

Test case $3$: Consider the following sequence of fights:
- Fight $1$: Bob and Claudio fight. Thus, the health level of both players reduces by $\texttt{min(1, 4)} = 1$. The final health levels of Bob and Claudio after the fight are $0$ and $3$ respectively.
- Fight $2$: Anna and Bob fight. Thus, the health level of both players reduces by $\texttt{min(5, 0)} = 0$. The final health levels of Anna and Bob after the fight are $5$ and $0$ respectively.
- Fight $3$: Anna and Claudio fight. Thus, the health level of both players reduces by $\texttt{min(5, 3)} = 3$. The final health levels of Anna and Claudio after the fight are $2$ and $0$ respectively.

Thus, in the above sequence of fights, Anna has a positive health remaining at the end.

Difficulty: EASY
Tags: Mathematics

Starter code:

Solutions:
t = int(input())
for i in range(0, t):
	(a, b, c) = map(int, input().split())
	x = min(b, c)
	b = b - x
	c = c - x
	if a > max(b, c):
		print('YES')
	else:
		print('NO')


for i in range(int(input())):
	(a, b, c) = map(int, input().split())
	if a - abs(b - c) > 0:
		print('YES')
	else:
		print('NO')


T = int(input())
for i in range(T):
	(A, B, C) = list(map(int, input().split()))
	healthLvl = min(B, C)
	B = B - healthLvl
	C = C - healthLvl
	Sum = B + C
	if A != 0 and A > Sum:
		print('YES')
	else:
		print('NO')


for i in range(int(input())):
	(x, y, z) = map(int, input().split())
	m = min(y, z)
	y -= m
	z -= m
	if x > y and x > z:
		print('YES')
	else:
		print('NO')


for i in range(int(input())):
	(a, b, c) = map(int, input().split())
	m = min(b, c)
	b -= m
	c -= m
	if a > max(b, c) and a != 0:
		print('YES')
	else:
		print('NO')


t = int(input())
for i in range(t):
	(x, y, z) = map(int, input().split())
	p = x
	q = y
	r = z
	a = min(x, y)
	x = x - a
	y = y - a
	b = min(y, z)
	y = y - b
	z = z - b
	c = min(x, z)
	x = x - c
	z = z - c
	if x > 0:
		print('YES')
	else:
		a = min(q, r)
		q = q - a
		r = r - a
		b = min(p, q)
		p = p - b
		q = q - b
		c = min(p, r)
		p = p - c
		r = r - c
		if p > 0:
			print('YES')
		else:
			print('NO')


for _ in range(int(input())):
	(a, b, c) = map(int, input().split())
	min3 = min(b, c)
	p = b - min3
	q = c - min3
	sum = p + q
	if a != 0 and a > sum:
		print('YES')
	else:
		print('NO')


for _ in range(int(input())):
	(a, b, c) = map(int, input().split())
	if a - (max(b, c) - min(b, c)) > 0 or b + c < a:
		print('YES')
	else:
		print('NO')


for i in range(int(input())):
	(x, y, z) = map(int, input().split())
	min3 = min(y, z)
	y = y - min3
	z = z - min3
	min4 = min(x, y)
	x = x - min4
	y = y - min4
	min5 = min(x, z)
	x = x - min5
	z = z - min5
	if x > 0:
		print('YES')
	else:
		print('NO')


for _ in range(int(input())):
	(x, y, z) = map(int, input().split())
	l = min(y, z)
	y -= l
	z -= l
	print('YES' if x > y + z else 'NO')


for i in range(0, int(input())):
	(a, b, c) = map(int, input().split())
	minimum = min(b, c)
	p = b - minimum
	q = c - minimum
	sum = p + q
	if a != 0 and a > sum:
		print('YES')
	else:
		print('NO')


n = int(input())
while n != 0:
	(num1, num2, num3) = map(int, input().split())
	if num1 > abs(num2 - num3):
		print('YES')
		n = n - 1
	else:
		print('NO')
		n = n - 1