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. 
#include <bits/stdc++.h>
using namespace std;
const long long MXN = 2e3 + 10;
const long long MXM = 5e2 + 10;
const long long MX5 = 1e5 + 10;
const long long MX6 = 1e6 + 10;
const long long LOG = 20;
const long long INF = 8e18;
const double eps = 1e-9;
const long long MOD = 1e9 + 7;
long long power(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2, md) % md
                      : power(a * a % md, b / 2, md) % md));
}
long long bmm(long long a, long long b) {
  return (a % b == 0 ? b : bmm(b, a % b));
}
string base2(long long n) {
  string a = "";
  while (n >= 2) {
    a += (char)(n % 2 + '0');
    n /= 2;
  }
  a += (char)(n + '0');
  reverse((a).begin(), (a).end());
  return a;
}
void solve() {
  long long x;
  string s;
  cin >> x >> s;
  long long ans = s.size();
  for (int i = 0; i < x; i++) {
    long long now = s[i] - '0', r = s.size() - i - 1;
    if (now == 1) continue;
    if (s.size() < x) {
      string Now = s.substr(i + 1, r);
      for (int j = 0; j < now - 1; j++) {
        s += Now;
      }
    }
    now--;
    ans = (ans + ((ans - i - 1 + MOD) * now) % MOD) % MOD;
  }
  cout << ans << '\n';
  return;
}
int main() {
  long long q;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}
