Problem:
Score : 600 points

Problem Statement

Snuke has a blackboard and a set S consisting of N integers.
The i-th element in S is S_i.

He wrote an integer X on the blackboard, then performed the following operation N times:

Choose one element from S and remove it.

Let x be the number written on the blackboard now, and y be the integer removed from S. Replace the number on the blackboard with x \bmod {y}.

There are N! possible orders in which the elements are removed from S.
For each of them, find the number that would be written on the blackboard after the N operations, and compute the sum of all those N! numbers modulo 10^{9}+7.

Constraints

All values in input are integers.

1 \leq N \leq 200

1 \leq S_i, X \leq 10^{5}

S_i are pairwise distinct.

Input

Input is given from Standard Input in the following format:

N X
S_1 S_2 \ldots S_{N}

Output

Print the answer.

Sample Input 1

2 19
3 7

Sample Output 1

3

There are two possible orders in which we remove the numbers from S.

If we remove 3 and 7 in this order, the number on the blackboard changes as follows: 19 \rightarrow 1 \rightarrow 1.

If we remove 7 and 3 in this order, the number on the blackboard changes as follows: 19 \rightarrow 5 \rightarrow 2.

The output should be the sum of these: 3.

Sample Input 2

5 82
22 11 6 5 13

Sample Output 2

288

Sample Input 3

10 100000
50000 50001 50002 50003 50004 50005 50006 50007 50008 50009

Sample Output 3

279669259

Be sure to compute the sum modulo 10^{9}+7.

Solution in JavaScript (node.js v5.12) (judge verdict: Accepted)

var GET=(function(){function f(s){return new g(s);}function g(s){this._s=s.trim().split("\n");this._y=0;}g.prototype.a=function(f){var s=this._s, y=this._y, r;if(typeof s[y]==="string")s[y]=s[y].split(" ").reverse();r=s[y].pop();if(!s[y].length)this._y++;return f?r:+r;};g.prototype.l=function(f){var s=this._s[this._y++].split(" ");return f?s:s.map(a=>+a);};g.prototype.m=function(n,f){var r=this._s.slice(this._y,this._y+=n).map(a=>a.split(" "));return f?r:r.map(a=>a.map(a=>+a));};g.prototype.r=function(n,f){var r=this._s.slice(this._y,this._y+=n);return f?r:r.map(a=>+a);};return f;})();
var o=GET(require("fs").readFileSync("/dev/stdin","utf8"));
function xArray(v){var a=arguments,l=a.length,r="Array(a["+--l+"]).fill(v)";while(--l)r="Array(a["+l+"]).fill(0).map(x=>"+r+")";return eval(r);}
var mod = 1e9+7;
function abmod(){for(var a=arguments,r=a[0],i=1;i<a.length;)r=((r>>16)*a[i]%mod*65536+(r&65535)*a[i++])%mod;return r;}

console.log(main());
function main(){
  var n = o.a();
  var x = o.a();
  var s = o.l().sort((a,b)=>b-a);
  var dp = Array(x+1).fill(0);
  dp[x] = 1;
  for(var i = 0; i < n; i++){
    var dp2 = Array(x+1).fill(0);
    for(var j = 0; j <= x; j++){
      dp2[j] += dp[j] * (n-i-1);
      dp2[j] %= mod;
      var m = j % s[i];
      dp2[m] += dp[j];
      dp2[m] %= mod;
    }
    dp = dp2;
  }
  var ans = 0;
  for(i = 0; i < s[n-1]; i++){
    ans += abmod(i,dp[i]);
  }
  return ans % mod;
}