Find nth root of m

"""
We need to find an integer x such that x^n = m (because nth root of m means raising the root to power n gives back m).
Since x^n increases as x increases, we can binary search x in the range [0, m].
For each mid, compute mid^n using pow(mid, n) and compare it with m.
If mid^n == m return mid, otherwise adjust the search space.
If no such integer exists, return -1.
"""

class Solution:
    def nthRoot(self, n, m):
        l,r=0,m
        while l<=r:
            mid=(l+r)//2
            val=pow(mid,n)
            if val==m:
                return mid
            elif val>m:
                r=mid-1
            else:
                l=mid+1
        return -1