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.io.*;

public class c 
{
	public static void main(String[] args) { new c(); }
	FS in = new FS();
	PrintWriter out = new PrintWriter(System.out);
	
	int t;
	int l, x, idx;
	char[] s, big;
	long ans, mod = (long)1e9 + 7;
	
	c() {
		t = in.nextInt();
		while (t --> 0) {
			x = in.nextInt();
			s = in.next().toCharArray();
			// filling big will only take O(x)
			// and sum x <= 10^6

			l = 0;
			idx = s.length;
			ans = s.length;
			big = new char[max(x, idx)];
			for (int i = 0; i < ans; i++)
				big[i] = s[i];
			while (l < x) {
				int to = (int)min(idx + (big[l] - '1') * (ans - (l + 1)), x);
				for (int i = 0; i < (big[l] - '1'); i++)
					for (int j = l + 1; idx < to && j < ans; j++)
						big[idx++] = big[j];
				ans = (l + 1) + (big[l] - '0') * (ans - (l + 1));
				ans = ((ans % mod) + mod) % mod;
				l++;
			}
			out.println(ans);
		}
		out.close();
	}
	
	

	int min(int a, int b) { if (a < b) return a; return b; }	
	int max(int a, int b) { if (a > b) return a; return b; }	
	long min(long a, long b) { if (a < b) return a; return b; }	
	long max(long a, long b) { if (a > b) return a; return b; }	

	boolean z(int x) { if (x == 0) return true; return false; }
	boolean z(long x) { if (x == 0) return true; return false; }
	
	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch (Exception e) {}
			} return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next()); }
		long nextLong() { return Long.parseLong(next()); }
		double nextDouble() { return Double.parseDouble(next()); }
	}
}
