<p>Given a binary string <code>s</code>, return <em>the number of substrings with all characters</em> <code>1</code><em>&#39;s</em>. Since the answer may be too 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> s = &quot;0110111&quot;
<strong>Output:</strong> 9
<strong>Explanation:</strong> There are 9 substring in total with only 1&#39;s characters.
&quot;1&quot; -&gt; 5 times.
&quot;11&quot; -&gt; 3 times.
&quot;111&quot; -&gt; 1 time.</pre>

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

<pre>
<strong>Input:</strong> s = &quot;101&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> Substring &quot;1&quot; is shown 2 times in s.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;111111&quot;
<strong>Output:</strong> 21
<strong>Explanation:</strong> Each substring contains only 1&#39;s characters.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution {
public:
    int numSub(string s) {
        
vector <long long> consecutive_one;
  const int MOD = 1e9 + 7;
int count = 0;
    for(int i = 0 ; i < s.length(); i++ )
    {
        if(s[i] == '1')
        {
            count++;
        }
        if(s[i] == '0')
        {
            if(count > 0)
                consecutive_one.push_back(count);
            count = 0;
        }
    }

    if(count > 0)
        consecutive_one.push_back(count);


unsigned long long total_sum = 0 ;

for(int i = 0; i < consecutive_one.size(); i++ )
{
//cout << consecutive_one[i] << "\n";
    total_sum = (total_sum + ((consecutive_one[i]+1)*consecutive_one[i])/2) % MOD ;
    
    }


//cout << total_sum << "\n";
    return total_sum;
    }
};
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