<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>

class Solution {
    int mod = 1e9+7;
    int p[100005];
    int prev[100005];
    int nxt[100005];
    bool ok[100005];
    int get_prev(int x){
        return prev[x] == -1?x:prev[x]=get_prev(prev[x]);
    }
    int get_nxt(int x){
        return nxt[x] == -1?x:nxt[x]=get_nxt(nxt[x]);
    }
    // int mn[400005];
    long long sum[100005];
    long long sum_l[100005]; //a[i]*i;
    long long sum_r[100005]; // a[i]*(n-i+1);
    // int mnsum[400005];
    int n;
    int get_left_sum(int L, int R){
        L++; R++;
        return (sum_l[R] - sum_l[L-1] + mod - (sum[R]-sum[L-1]+mod)%mod * (L-1)%mod + mod)%mod;
    }
    int get_right_sum(int L, int R){
        L++; R++;
        // n-R+1        n-L+1
        return (sum_r[R] - sum_r[L-1] + mod - (sum[R]-sum[L-1] + mod)%mod* (n-R)%mod + mod)%mod;
    }
public:
    int totalStrength(vector<int>& a) {
        n= a.size();
        for(int i=1;i<=n;i++){
            sum[i] = (sum[i-1] + a[i-1])%mod;
        }
        for(int i=1;i<=n;i++){
            sum_l[i] = (sum_l[i-1] + a[i-1]*1LL*i%mod)%mod;
        }
        for(int i=1;i<=n;i++){
            sum_r[i] = (sum_r[i-1]+ a[i-1]*1LL*(n-i+1)%mod)%mod;
        }
        for(int i=0;i<n;i++){
            p[i] = i;
        }
        sort(p, p+n, [&](int x,int y){return a[x]>a[y];});
        memset(prev, -1, sizeof prev);
        memset(nxt, -1, sizeof nxt);
        long long ans=0;
        for(int i=0;i<n;i++){
            long long mn = a[p[i]];
            ok[p[i]] = true;
            if(p[i]>0 && ok[p[i]-1]){
                int x = p[i];
                while(x>0 && ok[x-1]){
                    prev[x] = get_prev(x-1);
                    x = prev[x];
                }
                prev[p[i]] = x;
            }
            if(p[i]<n-1 && ok[p[i]+1]){
                int x = p[i];
                while(x<n-1 && ok[x+1] ){
                    nxt[x] = get_nxt(x+1);
                    x = nxt[x];
                }
                nxt[p[i]] = x;
            }
            long long left_num = 1; // empty
            long long right_num = 1; // empty
            long long left_sum = 0, right_sum = 0;
            if(prev[p[i]] != -1){
                left_sum = get_left_sum(prev[p[i]], p[i]-1);
                left_num = p[i] - prev[p[i]] +1;
            }
            if(nxt[p[i]] != -1){
                right_sum = get_right_sum(p[i]+1, nxt[p[i]]);
                right_num = nxt[p[i]] - p[i] + 1;
            }
            long long total_sum = (left_sum*1LL*right_num%mod + right_sum*1LL*left_num%mod + a[p[i]]*1LL*(left_num*1LL*right_num%mod)%mod)%mod;
            ans = (ans+total_sum*a[p[i]]%mod)%mod;
        }
        return ans;
    }
};
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