#include <iostream>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <algorithm>	// require sort next_permutation count __gcd reverse etc.
#include <cstdlib>	// require abs exit atof atoi 
#include <cstdio>		// require scanf printf
#include <functional>
#include <numeric>	// require accumulate
#include <cmath>		// require fabs
#include <climits>
#include <limits>
#include <cfloat>
#include <iomanip>	// require setw
#include <sstream>	// require stringstream 
#include <cstring>	// require memset
#include <cctype>		// require tolower, toupper
#include <fstream>	// require freopen
#include <ctime>		// require srand
#define rep(i,n) for(int i=0;i<(n);i++)
#define ALL(A) A.begin(), A.end()
/*

	No.4 おもりと天秤

	dp 解
	dp[i][j]:= i 番目のおもりと使ったときの 合計のおもり j

	dp[0][0] := true;
	dp[i+1][j] := dp[i][j], dp[i+1][j+w[i]] := dp[i][j] 	

	sum(W) % 2 == 0 かつ
	dp[N][sum(W)/2] ? possible : impossible

	計算量 O(N*sum(W))
*/

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int MAX_N = 105;
const int MAX_W = 105*105;

bool dp[MAX_N][MAX_W];

int main()
{
	memset (dp, false, sizeof (dp ) );
	ios_base::sync_with_stdio(0);
	int N; cin >> N;
	vector<int> W(N, 0 );
	rep (i, N ) cin >> W[i];
	int sumW = accumulate (ALL (W ), 0 );

	dp[0][0] = true;
	for (int i = 0; i < N; i++ ){
		for (int j = 0; j < MAX_W; j++ ){
			if (dp[i][j] ){
				dp[i+1][j] |= true;		// i 番目のおもりを使わない
				dp[i+1][j+W[i]] |= true;	// i 番目のおもりを使う
			} // end if
		} // end for
	} // end for

	bool res = ((sumW % 2 == 0 ) && dp[N][sumW/2] );
	cout << (res ? "possible" : "impossible" ) << endl;

	return 0;
}