We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i. 

There is one cursor. The cursor's location ℓ is denoted by an integer in \{0, …, |s|\}, with the following meaning: 

  * If ℓ = 0, then the cursor is located before the first character of s. 
  * If ℓ = |s|, then the cursor is located right after the last character of s. 
  * If 0 < ℓ < |s|, then the cursor is located between s_ℓ and s_{ℓ+1}. 



We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor. 

We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions:

  * The Move action. Move the cursor one step to the right. This increments ℓ once. 
  * The Cut action. Set c ← s_right, then set s ← s_left. 
  * The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c. 



The cursor initially starts at ℓ = 0. Then, we perform the following procedure:

  1. Perform the Move action once. 
  2. Perform the Cut action once. 
  3. Perform the Paste action s_ℓ times. 
  4. If ℓ = x, stop. Otherwise, return to step 1. 



You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7. 

It is guaranteed that ℓ ≤ |s| at any time.

Input

The first line of input contains a single integer t (1 ≤ t ≤ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases.

The first line of each test case contains a single integer x (1 ≤ x ≤ 10^6). The second line of each test case consists of the initial string s (1 ≤ |s| ≤ 500). It is guaranteed, that s consists of the characters "1", "2", "3".

It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that ℓ ≤ |s| at any time.

Output

For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7. 

Example

Input


4
5
231
7
2323
6
333
24
133321333


Output


25
1438
1101
686531475

Note

Let's illustrate what happens with the first test case. Initially, we have s =  231. Initially, ℓ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above:

  * Step 1, Move once: we get ℓ = 1. 
  * Step 2, Cut once: we get s =  2 and c =  31. 
  * Step 3, Paste s_ℓ =  2 times: we get s =  23131. 
  * Step 4: ℓ = 1 not= x = 5, so we return to step 1. 

  * Step 1, Move once: we get ℓ = 2. 
  * Step 2, Cut once: we get s =  23 and c =  131. 
  * Step 3, Paste s_ℓ =  3 times: we get s =  23131131131. 
  * Step 4: ℓ = 2 not= x = 5, so we return to step 1. 

  * Step 1, Move once: we get ℓ = 3. 
  * Step 2, Cut once: we get s =  231 and c =  31131131. 
  * Step 3, Paste s_ℓ =  1 time: we get s =  23131131131. 
  * Step 4: ℓ = 3 not= x = 5, so we return to step 1. 

  * Step 1, Move once: we get ℓ = 4. 
  * Step 2, Cut once: we get s =  2313 and c =  1131131. 
  * Step 3, Paste s_ℓ =  3 times: we get s =  2313113113111311311131131. 
  * Step 4: ℓ = 4 not= x = 5, so we return to step 1. 

  * Step 1, Move once: we get ℓ = 5. 
  * Step 2, Cut once: we get s =  23131 and c =  13113111311311131131. 
  * Step 3, Paste s_ℓ =  1 times: we get s =  2313113113111311311131131. 
  * Step 4: ℓ = 5 = x, so we stop. 



At the end of the procedure, s has length 25. 
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
  
    void solve() throws IOException {
        for (int TC=ni();TC>0;TC--) {
            long m=1000000007;
            int x=ni();
            String s=next();
            int[]A=new int[1000100];
            for (int z=0;z<s.length();z++) A[z+1]=(int)(s.charAt(z)-'0');
            
            long n=s.length();
            int p=0;
            int q=(int)n+1;
            
            while (q<=x) {
                p++;
                int c=A[p]-1;
                
                outer:
                for (int z=0;z<c;z++) {
                    for (int y=p+1;y<=n;y++) {
                        A[q]=A[y];
                        q++;
                        if (q>x) break outer;
                    }
                }
                n=n+(c*(n-p));
            }
            
            while (p<x) {
                p++;
                long r=(n+m-p)%m;
                n=(n+r*(A[p]-1))%m;
            }
            out.println(n);
        }
        out.flush();
    }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}