<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>

const int m = 1e9 + 7;
class Solution {
public:
    Solution(){
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
    }
    
    int numOfSubarrays(vector<int>& arr) {
        int even = arr[0]%2 == 0 ? 1 : 0;
        int odd = even == 1 ? 0 : 1;
        int ans = odd;

        for(int i=1;i<arr.size();i++){
            arr[i] = arr[i] + arr[i-1];
            if(arr[i]%2 == 0){
                ans = (ans + odd%m)%m;
                even++;
            }else{
                odd++;
                ans = (ans + even%m + 1)%m;
            }
        }

        return ans;
    }
};
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