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.io.*;
import java.util.*;
 
 
public class Main{
static int mod = (int)(Math.pow(10, 9) + 7);
   public static void main(String[] args) {
      MyScanner sc = new MyScanner();
      out = new PrintWriter(new BufferedOutputStream(System.out));
      int t=  sc.nextInt();
      while (t-->0) {
    	  int x = sc.nextInt();
          String s = sc.next();
          StringBuilder str = new StringBuilder(s);
          long hangoff = 0;
          int index = 0;
            while (str.length() < x && index < x){
                // do it manually
                int cur = str.charAt(index) - '0';
                if (cur == 1) {
                    index++;
                    continue;
                }
                StringBuilder temp = new StringBuilder(str.substring(index+1, str.length()));

                
                for (int j = 0; j < cur-1; j++){
                    str.append(temp);
                }
                index++;
            }
            if (index == x){
                out.println(str.length());
            }
            else{
                hangoff =  str.length()-index;
               // out.println(str + " " + str.length() + " " + hangoff);
                                
                for (int i = index; i < x; i++){
                    hangoff--;
                    long cur = str.charAt(i) - '0';
                    hangoff = (cur) * (hangoff);
                    hangoff %= mod;
                    //out.println(hangoff + " " + cur + " ");
                }
                out.println((hangoff + Integer.min(x, str.length())) % mod);
            }
      }
      
      out.close();
   }
   
   static long pow(long a, long N) {
	   if (N == 0) return 1;
	   else if (N == 1) return a;
	   else {
		   long R = pow(a,N/2);
		   if (N % 2 == 0) {
			   return R*R;
		   }
		   else {
			   return R*R*a;
		   }
	   }
   }
   
   static long powMod(long a, long N) {
	   if (N == 0) return 1;
	   else if (N == 1) return a % mod;
	   else {
		   long R = powMod(a,N/2) % mod;
		   R *= R % mod;
		   if (N % 2 == 1) {
			   R *= a % mod;
		   }
			   return R % mod;
	   }
   }
static void mergeSort(int[] A){ // low to hi sort, single array only
      int n = A.length;
      if (n < 2) return;
      int[] l = new int[n/2];
      int[] r = new int[n - n/2];
 
      for (int i = 0; i < n/2; i++){
        l[i] = A[i];
      }
 
      for (int j = n/2; j < n; j++){
        r[j-n/2] = A[j];
      }
 
      mergeSort(l);
      mergeSort(r);
      merge(l, r, A);
    }
 
    static void merge(int[] l, int[] r, int[] a){
      int i = 0, j = 0, k = 0;
      while (i < l.length && j < r.length && k < a.length){
        if (l[i] < r[j]){
          a[k] = l[i];
          i++;
        }
        else{
          a[k] = r[j];
          j++;
        }
        k++;
      }
      while (i < l.length){
        a[k] = l[i];
        i++;
        k++;
      }
 
      while (j < r.length){
        a[k] = r[j];
        j++;
        k++;
      }
    }

   

   //-----------PrintWriter for faster output---------------------------------
   public static PrintWriter out;
      
   //-----------MyScanner class for faster input----------
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;
 
      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }
 
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
 
      int nextInt() {
          return Integer.parseInt(next());
      }
 
      long nextLong() {
          return Long.parseLong(next());
      }
 
      double nextDouble() {
          return Double.parseDouble(next());
      }
 
      String nextLine(){
          String str = "";
	  try {
	     str = br.readLine();
	  } catch (IOException e) {
	     e.printStackTrace();
	  }
	  return str;
      }

   }
   //--------------------------------------------------------
}





