<p>You are given an <code>m x n</code> integer matrix <code>points</code> (<strong>0-indexed</strong>). Starting with <code>0</code> points, you want to <strong>maximize</strong> the number of points you can get from the matrix.</p>

<p>To gain points, you must pick one cell in <strong>each row</strong>. Picking the cell at coordinates <code>(r, c)</code> will <strong>add</strong> <code>points[r][c]</code> to your score.</p>

<p>However, you will lose points if you pick a cell too far from the cell that you picked in the previous row. For every two adjacent rows <code>r</code> and <code>r + 1</code> (where <code>0 &lt;= r &lt; m - 1</code>), picking cells at coordinates <code>(r, c<sub>1</sub>)</code> and <code>(r + 1, c<sub>2</sub>)</code> will <strong>subtract</strong> <code>abs(c<sub>1</sub> - c<sub>2</sub>)</code> from your score.</p>

<p>Return <em>the <strong>maximum</strong> number of points you can achieve</em>.</p>

<p><code>abs(x)</code> is defined as:</p>

<ul>
	<li><code>x</code> for <code>x &gt;= 0</code>.</li>
	<li><code>-x</code> for <code>x &lt; 0</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong><strong> </strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/12/screenshot-2021-07-12-at-13-40-26-diagram-drawio-diagrams-net.png" style="width: 300px; height: 300px;" />
<pre>
<strong>Input:</strong> points = [[1,2,3],[1,5,1],[3,1,1]]
<strong>Output:</strong> 9
<strong>Explanation:</strong>
The blue cells denote the optimal cells to pick, which have coordinates (0, 2), (1, 1), and (2, 0).
You add 3 + 5 + 3 = 11 to your score.
However, you must subtract abs(2 - 1) + abs(1 - 0) = 2 from your score.
Your final score is 11 - 2 = 9.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/12/screenshot-2021-07-12-at-13-42-14-diagram-drawio-diagrams-net.png" style="width: 200px; height: 299px;" />
<pre>
<strong>Input:</strong> points = [[1,5],[2,3],[4,2]]
<strong>Output:</strong> 11
<strong>Explanation:</strong>
The blue cells denote the optimal cells to pick, which have coordinates (0, 1), (1, 1), and (2, 0).
You add 5 + 3 + 4 = 12 to your score.
However, you must subtract abs(1 - 1) + abs(1 - 0) = 1 from your score.
Your final score is 12 - 1 = 11.
</pre>

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

<ul>
	<li><code>m == points.length</code></li>
	<li><code>n == points[r].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= points[r][c] &lt;= 10<sup>5</sup></code></li>
</ul>

bool y() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    ofstream out("user.out");
    long long leftMax[100000];
    string s;
    while(getline(cin,s)) {
        // parse s into vector<vector<int>>
        int i = 2, M = 1, N = 0, SL = (int) s.length();
        long long ans = 0;
        vector<vector<long long>> points;
        {
            long long curr = 0;
            points.push_back({});
            while(s[i] != ']') {
                if('0' <= s[i] && s[i] <= '9') {
                    curr = 10*curr+(s[i]-'0');
                }else {
                    points[0].push_back(curr);
                    if(curr > ans) ans = curr;
                    curr = 0;
                    N++;
                }
                i++;
            }
            points[0].push_back(curr);
            N++;
            if(curr > ans) ans = curr;
            i += 3;
        }
        while(i < SL) {
            int X = 0;
            long long curr = 0;
            points.push_back(vector<long long>(N));
            while(s[i] != ']') {
                if('0' <= s[i] && s[i] <= '9') {
                    curr = 10*curr+(s[i]-'0');
                }else {
                    points[M][X++] = curr;
                    curr = 0;
                }
                i++;
            }
            points[M][X] = curr;
            i += 3;
            M++;
        }
        // maxPoints()
        for(int i = 1; i < M; i++) { // points[0] acts as dp
            leftMax[0] = points[0][0];
            for(int j = 1; j < N; j++) {
                leftMax[j] = max(leftMax[j-1] - 1, points[0][j]);
            }
            long long prevRightMax = points[0][N-1];
            ans = max(ans, points[0][N-1] = points[i][N-1] + max(leftMax[N-1], prevRightMax));
            for(int j = N-1; j --> 0;) {
                prevRightMax = max(prevRightMax - 1, points[0][j]);
                ans = max(ans, points[0][j] = points[i][j] + max(leftMax[j], prevRightMax));
            }
        }
        out << ans << '\n';
    }
    out.flush();
    exit(0);
    return 'a';
};
bool x = y();
struct Solution {
    long long maxPoints(const vector<vector<int>>& points) {return 0ll;}
};
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