<p>You are given two <strong>sorted</strong> arrays of distinct integers <code>nums1</code> and <code>nums2</code>.</p>

<p>A <strong>valid<strong><em> </em></strong>path</strong> is defined as follows:</p>

<ul>
	<li>Choose array <code>nums1</code> or <code>nums2</code> to traverse (from index-0).</li>
	<li>Traverse the current array from left to right.</li>
	<li>If you are reading any value that is present in <code>nums1</code> and <code>nums2</code> you are allowed to change your path to the other array. (Only one repeated value is considered in the valid path).</li>
</ul>

<p>The <strong>score</strong> is defined as the sum of unique values in a valid path.</p>

<p>Return <em>the maximum score you can obtain of all possible <strong>valid paths</strong></em>. Since the answer may be too large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/16/sample_1_1893.png" style="width: 500px; height: 151px;" />
<pre>
<strong>Input:</strong> nums1 = [2,4,5,8,10], nums2 = [4,6,8,9]
<strong>Output:</strong> 30
<strong>Explanation:</strong> Valid paths:
[2,4,5,8,10], [2,4,5,8,9], [2,4,6,8,9], [2,4,6,8,10],  (starting from nums1)
[4,6,8,9], [4,5,8,10], [4,5,8,9], [4,6,8,10]    (starting from nums2)
The maximum is obtained with the path in green <strong>[2,4,6,8,10]</strong>.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [1,3,5,7,9], nums2 = [3,5,100]
<strong>Output:</strong> 109
<strong>Explanation:</strong> Maximum sum is obtained with the path <strong>[1,3,5,100]</strong>.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [1,2,3,4,5], nums2 = [6,7,8,9,10]
<strong>Output:</strong> 40
<strong>Explanation:</strong> There are no common elements between nums1 and nums2.
Maximum sum is obtained with the path [6,7,8,9,10].
</pre>

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

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 10<sup>7</sup></code></li>
	<li><code>nums1</code> and <code>nums2</code> are strictly increasing.</li>
</ul>

// Bài này hiểm ở đoạn xử lý số lớn
void printArr(const vector<long long> & a){
    for(int i = 0 ; i < a.size() ; i ++) cout<<a[i]<<" ";
    cout<<endl;
}
int BinarySearch(const vector<int> & a, int val){
    int l = 0 , r = a.size() - 1;
    int idx = -1;
    while(l <= r){
        int m = l + (r - l) / 2;
        if(a[m] == val){
            return m;
        }else if(a[m] < val){
            l = m + 1;
        }else if(a[m] > val){
            r = m - 1;
        }
    }
    return idx;
}
class Solution {
public:
    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        long long mod = 1e9 + 7;
        unordered_set<long long> s;
        int i;
        for(i = 0 ; i < nums1.size() ; i ++) s.insert(nums1[i]);
        for(i = 0 ; i < nums2.size() ; i ++) s.insert(nums2[i]);
        vector<long long> a;
        for(auto ele = s.begin(); ele != s.end(); ele ++) a.push_back(*ele);
        sort(a.begin(), a.end());
        vector<long long> a1 = a, a2 = a;
        for(i = 0 ; i < a1.size() ; i ++){
            int idx = BinarySearch(nums1, (int)a1[i]);
            if(idx == -1) a1[i] = 0;
        }
        for(i = 0 ; i < a2.size() ; i ++){
            int idx = BinarySearch(nums2, (int)a2[i]);
            if(idx == -1) a2[i] = 0;
        }
        //printArr(a1);
        //printArr(a2);
        for(i = 1 ; i < a1.size() ; i ++){
            long long tong1 = a1[i - 1] + a1[i];
            long long tong2 = a2[i - 1] + a2[i];
            if(a1[i] == a2[i]){
                a1[i] = max(tong1, tong2);
                a2[i] = max(tong1, tong2);
            }else{
                a1[i] = tong1;
                a2[i] = tong2;
            }
        }
        //cout<<max(a1.back(), a2.back());
        return (int) (max(a1.back(), a2.back()) % mod);
    }
};
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