<p>You are given a 2D array of axis-aligned <code>rectangles</code>. Each <code>rectangle[i] = [x<sub>i1</sub>, y<sub>i1</sub>, x<sub>i2</sub>, y<sub>i2</sub>]</code> denotes the <code>i<sup>th</sup></code> rectangle where <code>(x<sub>i1</sub>, y<sub>i1</sub>)</code> are the coordinates of the <strong>bottom-left corner</strong>, and <code>(x<sub>i2</sub>, y<sub>i2</sub>)</code> are the coordinates of the <strong>top-right corner</strong>.</p>

<p>Calculate the <strong>total area</strong> covered by all <code>rectangles</code> in the plane. Any area covered by two or more rectangles should only be counted <strong>once</strong>.</p>

<p>Return <em>the <strong>total area</strong></em>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/06/rectangle_area_ii_pic.png" style="width: 600px; height: 450px;" />
<pre>
<strong>Input:</strong> rectangles = [[0,0,2,2],[1,0,2,3],[1,0,3,1]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> A total area of 6 is covered by all three rectangles, as illustrated in the picture.
From (1,1) to (2,2), the green and red rectangles overlap.
From (1,0) to (2,3), all three rectangles overlap.
</pre>

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

<pre>
<strong>Input:</strong> rectangles = [[0,0,1000000000,1000000000]]
<strong>Output:</strong> 49
<strong>Explanation:</strong> The answer is 10<sup>18</sup> modulo (10<sup>9</sup> + 7), which is 49.
</pre>

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

<ul>
	<li><code>1 &lt;= rectangles.length &lt;= 200</code></li>
	<li><code>rectanges[i].length == 4</code></li>
	<li><code>0 &lt;= x<sub>i1</sub>, y<sub>i1</sub>, x<sub>i2</sub>, y<sub>i2</sub> &lt;= 10<sup>9</sup></code></li>
	<li><code>x<sub>i1 &lt;= </sub>x<sub>i2</sub></code></li>
	<li><code>y<sub>i1 &lt;=</sub> y<sub>i2</sub></code></li>
</ul>

bool compare(vector<long long> v1, vector<long long> v2){
    if(v1[0] < v2[0]){
        return true;
    }
    else if(v1[0] > v2[0]){
        return false;
    }
    if(v1[1] <= v2[1]){
        return true;
    }
    return false;
}



class Solution {
public:
    map<pair<long long,long long>, long long> m;
    long long mod = 1e9+7;
    long long find(long long diff){
        if(m.size() == 0){
            return 0;
        }
        long long start = -1;
        long long last = -1;
        long long total = 0;
        for(auto a:m){
            // cout<<a.first.first<<"a "<<a.first.second<<endl;
            if(start == -1){
                start = a.first.first;
                last = a.first.second;
            }
            else{
                if(last < a.first.first){
                    total += last - start;
                    start = a.first.first;
                    last = a.first.second;
                    // cout<<"b";
                }
                else{
                    last = max(last, a.first.second);
                    // cout<<"c";
                }
            }
        }
        total += last - start;
        // cout<<total<<" "<<diff<<endl<<endl<<endl;
        long long ret = (total*diff)%mod;
        if(ret < (long long)0){
            ret += mod;
        }
        return ret;
    }
    int rectangleArea(vector<vector<int>>& rectangles) {
        vector<vector<long long>> events;
        for(auto a:rectangles){
            events.push_back({(long long)a[0], (long long)0, (long long)a[1], (long long)a[3]});
            events.push_back({(long long)a[2], (long long)1, (long long)a[1], (long long)a[3]});
        }
        sort(events.begin(), events.end(), compare);
        // for(auto a:events){
        //     cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<" "<<a[3]<<endl;
        // }
        long long ans = 0;
        long long n = events.size();
        for(long long i=0;i<n-1;i++){
            if(events[i][1] == 0){
                m[{events[i][2], events[i][3]}]++;
            }
            else{
                m[{events[i][2], events[i][3]}]--;
                if(m[{events[i][2], events[i][3]}] == 0){
                    m.erase({events[i][2], events[i][3]});
                }
            }
            // cout<<i<<" "<<endl;
            ans = (ans + find(events[i+1][0] - events[i][0]))%mod;
            if(ans < 0){
                ans += mod;
            }
        }
        return ans;
    }
};
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