<p>Given an array of integers <code>arr</code>, return <em>the number of subarrays with an <strong>odd</strong> sum</em>.</p>

<p>Since the answer can be very large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [1,3,5]
<strong>Output:</strong> 4
<strong>Explanation:</strong> All subarrays are [[1],[1,3],[1,3,5],[3],[3,5],[5]]
All sub-arrays sum are [1,4,9,3,8,5].
Odd sums are [1,9,3,5] so the answer is 4.
</pre>

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

<pre>
<strong>Input:</strong> arr = [2,4,6]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All subarrays are [[2],[2,4],[2,4,6],[4],[4,6],[6]]
All sub-arrays sum are [2,6,12,4,10,6].
All sub-arrays have even sum and the answer is 0.
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,2,3,4,5,6,7]
<strong>Output:</strong> 16
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= 100</code></li>
</ul>

class Solution {
public:
    int numOfSubarrays(vector<int>& arr) {
        int sum = 0;
        //  1 4 9
        //  1 2 3  4  5  6  7
        //  1 3 6 10 15 21 28
        //  +1+1

        // even + odd = odd;
        // even + even =even 


        // 100 100 99 99 
        // even++ 
        // 100 200 299 398
        int n = arr.size();
        vector<int> prefix(n);
        prefix[0] = arr[0];
        long long int cnt =0,odd = 0,even =0;
        int mod = 1e9+7;
        for(int i=1;i<arr.size();i++)
        {
            prefix[i] = prefix[i-1] + arr[i];
        }

        for(int i=0;i<n;i++)
        {
            if(prefix[i]%2==0)
            {
                even++;
            }
            else odd++;
        }
        return (odd+(odd*even)%mod)%mod;
    }
};
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