<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) {
        int mod= (1e9+7);
        long long count=0;
        int f=0,b=0;
        stack<int> st;
        while(b<s.size() && f<s.size()){
            if(s[f]=='1'){
                int c=1;
                b=f+1;
                while(b!=s.size() && s[b]=='1'){
                    c++;
                    b++;
                }
                st.push(c);
            }
            f=b+1;
            b++;
        }
        while(!st.empty()){
            int t=st.top();
            long long temp = static_cast<long long>(t) * (t + 1) / 2;
            count = (count + temp % mod) % mod;
            st.pop();
            
        }
        return count;
    }
};
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