Raju has created a program to find the square root of a number. But his program can store only integers. Being a newbie, he didn't know about rounding the numbers. Hence his program returns the absolute value of the result if possible. For example, sqrt(3) = 1.73205080757……. His program will return 1
Given a number $N$, and it's integral square root $S$, His instructor will consider the answer correct if Difference between $N$ and the square of $S$ is within less than or equal to $X$% of $N$.
-----Input:-----
- First line contains $T$ no. of test cases and $X$ separated by space
- For every test case, a line contains an integer $N$
-----Output:-----
For every test case, print yes  if his programs return square root and (N-(S^2)) <= 0.01XN . For everything else, print no on a new line
-----Constraints-----
10 points:
- $1 \leq T \leq 10$
- $0\leq N \leq 10$
20 points:
- $1 \leq T \leq 30000$
- $-10^9 \leq N \leq 10^9$
70 points:
- $1 \leq T \leq 10^6$
- $-10^9 \leq N \leq 10^9$
-----Sample Input:-----
2 20
5
3
-----Sample Output:-----
yes
no
-----EXPLANATION:-----
In #1, sqrt(5) = 2.2360679775. Taking integral value, S = 2.
S2 = 4. Difference=1 which is within 20% of 5
In #1, sqrt(3) = 1.73205080757. Taking integral value, S = 1.
S2 = 1. Difference=2  which is not within 20% of 3
# cook your dish here
import math as m
tst,N = list(map(int,input().split()))
for i in range(tst):
 val = int(input())
 #z = m.floor(z)
 s = int(m.sqrt(val))
 S = pow(s,2)
 p = val-S
 q = int((N*val)*(.01))
 if p<=q:
  print("yes")
 else:
  print("no")


# cook your dish here
import math as m
tst,N = list(map(int,input().split()))
for i in range(tst):
 val = int(input())
 #z = m.floor(z)
 s = int(m.sqrt(val))
 S = pow(s,2)
 p = val-S
 q = int((N*val)*(.01))
 if p<=q:
  print("yes")
 else:
  print("no")

