##### Short Problem Definition:

Shashank loves to play with arrays a lot. Today, he has an array A consisting of N positive integers. At first, Shashank listed all the subarrays of his array A on a paper and later replaced all the subarrays on the paper with the maximum element present in the respective subarray.

##### Link

##### Complexity:

time complexity is `O(N)`

space complexity is `O(1)`

##### Execution:

You do not actually need to construct all the sub-arrays, as they reduce to only one element. You also can ignore all sub-arrays, that do not contain elements E > K. I also observed that there are x*y sub-arrays that match the above specified criteria for each element E > K. X is the distance from E to any previous e > K. Y is the distance from E to the end of the array. This way you crate all the sub-arrays that contain E and are not part of another e.

##### Solution:

```
#!/usr/bin/py
def numberList(a ,k):
result = 0
last_biggest = -1
a_len = len(a)
for idx in xrange(a_len):
if a[idx] > k:
result += (idx-last_biggest)*(a_len-idx)
last_biggest = idx
return result
if __name__ == '__main__':
t = int(raw_input())
for _ in xrange(t):
n,k = map(int, raw_input().split())
a = map(int, raw_input().split())
print numberList(a ,k)
```