# 1680. Concatenation of Consecutive Binary Numbers

#### Medium

Given an integer `n`, return *the **decimal value** of the binary string formed by concatenating the binary representations of* `1` *to* `n` *in order, **modulo*** `109 + 7`.

 

**Example 1:**

<pre><code>Input: n = 1
<strong>Output:
</strong> 1
<strong>Explanation: 
</strong>"1" in binary corresponds to the decimal value 1. 
</code></pre>

**Example 2:**

<pre><code>Input: n = 3
<strong>Output:
</strong> 27
<strong>Explanation: 
</strong>In binary, 1, 2, and 3 corresponds to "1", "10", and "11".
After concatenating them, we have "11011", which corresponds to the decimal value 27.
</code></pre>

**Example 3:**

<pre><code>Input: n = 12
<strong>Output:
</strong> 505379714
<strong>Explanation
</strong>: The concatenation results in "1101110010111011110001001101010111100".
The decimal value of that is 118505380540.
After modulo 109 + 7, the result is 505379714.
</code></pre>

&#x20;

**Constraints:**

* `1 <= n <= 105`

```python
class Solution:
    def concatenatedBinary(self, n: int) -> int:
        result = ""
        MOD = 10**9+7
        for num in range(1, n+1):
            b = bin(num)[2:]
            result += b
        return int(result, 2) % MOD
```