Given a **0-indexed** integer array `nums`, return _the **smallest** index_ `i` _of_ `nums` _such that_ `i mod 10 == nums[i]`_, or_ `-1` _if such index does not exist_.

`x mod y` denotes the **remainder** when `x` is divided by `y`.

**Example 1:**

**Input:** nums = \[0,1,2\]
**Output:** 0
**Explanation:** 
i=0: 0 mod 10 = 0 == nums\[0\].
i=1: 1 mod 10 = 1 == nums\[1\].
i=2: 2 mod 10 = 2 == nums\[2\].
All indices have i mod 10 == nums\[i\], so we return the smallest index 0.

**Example 2:**

**Input:** nums = \[4,3,2,1\]
**Output:** 2
**Explanation:** 
i=0: 0 mod 10 = 0 != nums\[0\].
i=1: 1 mod 10 = 1 != nums\[1\].
i=2: 2 mod 10 = 2 == nums\[2\].
i=3: 3 mod 10 = 3 != nums\[3\].
2 is the only index which has i mod 10 == nums\[i\].

**Example 3:**

**Input:** nums = \[1,2,3,4,5,6,7,8,9,0\]
**Output:** -1
**Explanation:** No index satisfies i mod 10 == nums\[i\].

**Constraints:**

*   `1 <= nums.length <= 100`
*   `0 <= nums[i] <= 9`


def smallest_index(nums):
    for i, num in enumerate(nums):
        if i % 10 == num:
            return i
    return -1
