<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>.</p>

<p>A <strong>subsequence</strong> of <code>nums</code> having length <code>k</code> and consisting of <strong>indices</strong> <code>i<sub>0</sub>&nbsp;&lt;&nbsp;i<sub>1</sub> &lt;&nbsp;... &lt; i<sub>k-1</sub></code> is <strong>balanced</strong> if the following holds:</p>

<ul>
	<li><code>nums[i<sub>j</sub>] - nums[i<sub>j-1</sub>] &gt;= i<sub>j</sub> - i<sub>j-1</sub></code>, for every <code>j</code> in the range <code>[1, k - 1]</code>.</li>
</ul>

<p>A <strong>subsequence</strong> of <code>nums</code> having length <code>1</code> is considered balanced.</p>

<p>Return <em>an integer denoting the <strong>maximum</strong> possible <strong>sum of elements</strong> in a <strong>balanced</strong> subsequence of </em><code>nums</code>.</p>

<p>A <strong>subsequence</strong> of an array is a new <strong>non-empty</strong> array that is formed from the original array by deleting some (<strong>possibly none</strong>) of the elements without disturbing the relative positions of the remaining elements.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,3,5,6]
<strong>Output:</strong> 14
<strong>Explanation:</strong> In this example, the subsequence [3,5,6] consisting of indices 0, 2, and 3 can be selected.
nums[2] - nums[0] &gt;= 2 - 0.
nums[3] - nums[2] &gt;= 3 - 2.
Hence, it is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.
The subsequence consisting of indices 1, 2, and 3 is also valid.
It can be shown that it is not possible to get a balanced subsequence with a sum greater than 14.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,-1,-3,8]
<strong>Output:</strong> 13
<strong>Explanation:</strong> In this example, the subsequence [5,8] consisting of indices 0 and 3 can be selected.
nums[3] - nums[0] &gt;= 3 - 0.
Hence, it is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.
It can be shown that it is not possible to get a balanced subsequence with a sum greater than 13.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [-2,-1]
<strong>Output:</strong> -1
<strong>Explanation:</strong> In this example, the subsequence [-1] can be selected.
It is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def maxBalancedSubsequenceSum(self, nums: List[int]) -> int:
        n = len(nums)
        dp = [0] * n
        values = [nums[i] - i for i in range(n)]
        sorted_values = sorted(set(values))
        compress_map = {val: idx for idx, val in enumerate(sorted_values)}
        seg_tree = [-float('inf')] * (4 * len(sorted_values))

        def query(l, r, st, en, node):
            if l > en or r < st:
                return -float('inf')
            if l <= st and en <= r:
                return seg_tree[node]
            md = (st + en) // 2
            return max(query(l, r, st, md, 2 * node), query(l, r, md + 1, en, 2 * node + 1))
        
        def update(ind, val, st, en, node):
            if st == en:
                seg_tree[node] = max(seg_tree[node], val)
                return
            md = (st + en) // 2
            if ind <= md:
                update(ind, val, st, md, 2 * node)
            else:
                update(ind, val, md + 1, en, 2 * node + 1)
            seg_tree[node] = max(seg_tree[2 * node], seg_tree[2 * node + 1])
        for i in range(n):
            compressed_index = compress_map[values[i]]
            max_prev_dp = query(0, compressed_index, 0, len(sorted_values) - 1, 1)
            dp[i] = nums[i]
            if max_prev_dp != -float('inf'):
                dp[i] = max(dp[i], nums[i] + max_prev_dp)
            update(compressed_index, dp[i], 0, len(sorted_values) - 1, 1)
        return max(dp)

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