<p>A split of an integer array is <strong>good</strong> if:</p>

<ul>
	<li>The array is split into three <strong>non-empty</strong> contiguous subarrays - named <code>left</code>, <code>mid</code>, <code>right</code> respectively from left to right.</li>
	<li>The sum of the elements in <code>left</code> is less than or equal to the sum of the elements in <code>mid</code>, and the sum of the elements in <code>mid</code> is less than or equal to the sum of the elements in <code>right</code>.</li>
</ul>

<p>Given <code>nums</code>, an array of <strong>non-negative</strong> integers, return <em>the number of <strong>good</strong> ways to split</em> <code>nums</code>. As the number 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>

<pre>
<strong>Input:</strong> nums = [1,1,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only good way to split nums is [1] [1] [1].</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,2,2,5,0]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are three good ways of splitting nums:
[1] [2] [2,2,5,0]
[1] [2,2] [2,5,0]
[1,2] [2,2] [5,0]
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,2,1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no good way to split nums.</pre>

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

<ul>
	<li><code>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

const int mod = 1e9 + 7;

int count_bet(vector<int> &a, int ind, int low, int high) {
    if(low > high) return 0;
    
    auto l_itr = lower_bound(a.begin(), a.begin() + ind, low);
    auto r_itr = upper_bound(a.begin(), a.begin() + ind, high);
    return (r_itr - l_itr);
}

class Solution {
public:
    int waysToSplit(vector<int>& a) {
        int n = a.size();
		
        auto pref = a;
        for(int i = 1; i < n; i++) {
            pref[i] += pref[i - 1];
        }
        
        int sum = pref.back();
        long long cnt = 0;
        
        int suf = 0;
        for(int i = n - 1; i >= 2; i--) {
            suf += a[i];
            auto in_bet = count_bet(pref, i - 1, sum - 2*suf, (sum - suf)/2);
            cnt = (cnt + in_bet) % mod;
        }
        
        return cnt;
    }
};
89000