<p>As the ruler of a kingdom, you have an army of wizards at your command.</p>

<p>You are given a <strong>0-indexed</strong> integer array <code>strength</code>, where <code>strength[i]</code> denotes the strength of the <code>i<sup>th</sup></code> wizard. For a <strong>contiguous</strong> group of wizards (i.e. the wizards&#39; strengths form a <strong>subarray</strong> of <code>strength</code>), the <strong>total strength</strong> is defined as the <strong>product</strong> of the following two values:</p>

<ul>
	<li>The strength of the <strong>weakest</strong> wizard in the group.</li>
	<li>The <strong>total</strong> of all the individual strengths of the wizards in the group.</li>
</ul>

<p>Return <em>the <strong>sum</strong> of the total strengths of <strong>all</strong> contiguous groups of wizards</em>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>A <strong>subarray</strong> is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>

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

<pre>
<strong>Input:</strong> strength = [1,3,1,2]
<strong>Output:</strong> 44
<strong>Explanation:</strong> The following are all the contiguous groups of wizards:
- [1] from [<u><strong>1</strong></u>,3,1,2] has a total strength of min([1]) * sum([1]) = 1 * 1 = 1
- [3] from [1,<u><strong>3</strong></u>,1,2] has a total strength of min([3]) * sum([3]) = 3 * 3 = 9
- [1] from [1,3,<u><strong>1</strong></u>,2] has a total strength of min([1]) * sum([1]) = 1 * 1 = 1
- [2] from [1,3,1,<u><strong>2</strong></u>] has a total strength of min([2]) * sum([2]) = 2 * 2 = 4
- [1,3] from [<u><strong>1,3</strong></u>,1,2] has a total strength of min([1,3]) * sum([1,3]) = 1 * 4 = 4
- [3,1] from [1,<u><strong>3,1</strong></u>,2] has a total strength of min([3,1]) * sum([3,1]) = 1 * 4 = 4
- [1,2] from [1,3,<u><strong>1,2</strong></u>] has a total strength of min([1,2]) * sum([1,2]) = 1 * 3 = 3
- [1,3,1] from [<u><strong>1,3,1</strong></u>,2] has a total strength of min([1,3,1]) * sum([1,3,1]) = 1 * 5 = 5
- [3,1,2] from [1,<u><strong>3,1,2</strong></u>] has a total strength of min([3,1,2]) * sum([3,1,2]) = 1 * 6 = 6
- [1,3,1,2] from [<u><strong>1,3,1,2</strong></u>] has a total strength of min([1,3,1,2]) * sum([1,3,1,2]) = 1 * 7 = 7
The sum of all the total strengths is 1 + 9 + 1 + 4 + 4 + 4 + 3 + 5 + 6 + 7 = 44.
</pre>

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

<pre>
<strong>Input:</strong> strength = [5,4,6]
<strong>Output:</strong> 213
<strong>Explanation:</strong> The following are all the contiguous groups of wizards: 
- [5] from [<u><strong>5</strong></u>,4,6] has a total strength of min([5]) * sum([5]) = 5 * 5 = 25
- [4] from [5,<u><strong>4</strong></u>,6] has a total strength of min([4]) * sum([4]) = 4 * 4 = 16
- [6] from [5,4,<u><strong>6</strong></u>] has a total strength of min([6]) * sum([6]) = 6 * 6 = 36
- [5,4] from [<u><strong>5,4</strong></u>,6] has a total strength of min([5,4]) * sum([5,4]) = 4 * 9 = 36
- [4,6] from [5,<u><strong>4,6</strong></u>] has a total strength of min([4,6]) * sum([4,6]) = 4 * 10 = 40
- [5,4,6] from [<u><strong>5,4,6</strong></u>] has a total strength of min([5,4,6]) * sum([5,4,6]) = 4 * 15 = 60
The sum of all the total strengths is 25 + 16 + 36 + 36 + 40 + 60 = 213.
</pre>

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

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

using LL = long long;
class Solution {
    LL M = 1e9+7;
public:
    int totalStrength(vector<int>& strength) 
    {
        int n = strength.size();
        vector<LL> presum(n+2, 0);
        vector<LL> presum2(n+2, 0);
        vector<int> arr = {0};
        for(auto x:strength)
            arr.push_back(x);
        for(int i=1;i<=n;i++)
        {
            presum[i] = (presum[i-1] + (LL)arr[i])%M;
            presum2[i] = (presum2[i-1] + (LL)arr[i]*i)%M;
        }
        stack<int> stk;

        vector<LL> prevSmaller(n+2,0);
        vector<LL> nextSmaller(n+2,n+1);

        for(int i=0;i<=n;i++)
        {
            while(!stk.empty() && arr[stk.top()] > arr[i])
            {
                nextSmaller[stk.top()] = i;
                stk.pop();
            }
            if(!stk.empty())
                prevSmaller[i] = stk.top();
            stk.push(i);
        }
        LL ret = 0;
        for (int i=1; i<=n; i++)
        {
            LL a = prevSmaller[i], b = nextSmaller[i];
            LL x = i-a, y = b-i;
            LL first = ((presum2[i-1] - presum2[a]) - (presum[i-1] - presum[a]) * a %M + M) % M;
            first = first * y % M;
            LL second = ((presum[b-1] - presum[i]) * (b-1+1) - (presum2[b-1] - presum2[i]) + M ) % M;
            second = second * x % M;
            LL mid = (LL)arr[i] * x * y % M;
            
            ret = (ret +(first + second + mid) * arr[i]) % M;  
        }
        return ret;



    }
};
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