Problem:
Score : 300 points

Problem Statement

There is a box containing N balls. The i-th ball has the integer A_i written on it.
Snuke can perform the following operation any number of times:

Take out two balls from the box. Then, return them to the box along with a new ball, on which the absolute difference of the integers written on the two balls is written.

Determine whether it is possible for Snuke to reach the state where the box contains a ball on which the integer K is written.

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq K \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 ... A_N

Output

If it is possible for Snuke to reach the state where the box contains a ball on which the integer K is written, print POSSIBLE; if it is not possible, print IMPOSSIBLE.

Sample Input 1

3 7
9 3 4

Sample Output 1

POSSIBLE

First, take out the two balls 9 and 4, and return them back along with a new ball, abs(9-4)=5.
Next, take out 3 and 5, and return them back along with abs(3-5)=2.
Finally, take out 9 and 2, and return them back along with abs(9-2)=7.
Now we have 7 in the box, and the answer is therefore POSSIBLE.

Sample Input 2

3 5
6 9 3

Sample Output 2

IMPOSSIBLE

No matter what we do, it is not possible to have 5 in the box. The answer is therefore IMPOSSIBLE.

Sample Input 3

4 11
11 3 7 15

Sample Output 3

POSSIBLE

The box already contains 11 before we do anything. The answer is therefore POSSIBLE.

Sample Input 4

5 12
10 2 8 6 4

Sample Output 4

IMPOSSIBLE

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
#define tr(container, it) \
    for (auto it = container.begin(); it != container.end(); it++)
#define scontains(c, x) ((c).find(x) != (c).end())   //O(log n)
#define contains(c, x) (find((c).begin(),(c).end(),x) != (c).end()) //O(n)
#define ill(_x)  ll _x;scanf("%lld",&_x);
#define idb(_x)  double _x;scanf("%lf",&_x);
#define pll pair<ll,ll>
#define mll map<ll,ll>
#define vll vector<ll>
#define sll set<ll>
#define vs vector<string>
#define in0(x, a, b)((x)>=a && (x)<=b    )
#define in1(x, a, b)((x)>a && (x)<b)
#define  rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define endl '\n'
#define int ll
const double PI = 3.14159265358979323846;
const int INF = 0x3f3f3f3f;
const int MOD = (int) (1e9) + 7;
const int MX = 777777 + 5;
const int mod = 1777777777;
const double EPS = 1e-10;
const ll MXP = 8999251775770639633;


int fac[MX+5];
int power(int x, int y, int p)
{
    int res = 1;

    x = x % p;

    while (y)
    {
        if (y & 1)
            res = (res*x) % p;
        y = y>>1;
        x = (x*x) % p;
    }
    return res;
}

// Returns n^(-1) mod p
int modInverse(int n, int p)
{
    return power(n, p-2, p);
}

int C(int n, int r, int p)
{
    if (r==0)
        return 1;

    return (fac[n]* modInverse(fac[r], p) % p *
            modInverse(fac[n-r], p) % p) % p;
}

void init(int mod){
    fac[0] = 1;
    for (int i=1 ; i<=MX; i++)
        fac[i] = fac[i-1]*i%mod;
}

int CC(int a, int b){
    int res = 1;
    rep(i,a+1,a-b+1)res *= i,res/=a-i+1;
//    rep(i,1,b+1)res /= i;
    return res;
}
void _() {
    int a,k;
    cin >> a >>k;
    if(a==1){
        int t;
        cin >> t;
        if(t!=k)cout << "IMPOSSIBLE\n" ;
    }
    else{
        int g,t;
        cin >> g;
        rep(i,0,a-1)cin >> t, g=__gcd(t,g);
        cout << (k%g?"IMPOSSIBLE":"POSSIBLE") << endl;
    }

};

#undef int
int main() {
#if __MINGW32__
    freopen("C:\\Users\\hiroo\\CLionProjects\\competitive\\Input.txt", "r", stdin);
    freopen("C:\\Users\\hiroo\\CLionProjects\\competitive\\Output.txt", "w", stdout);

#endif

    _();
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);


}