#include "bits/stdc++.h" using namespace std; typedef long long ll; typedef pair pii; typedef pair pll; #define FOR(i, s, e) for (ll(i) = (s); (i) < (e); (i)++) #define FORR(i, s, e) for (ll(i) = (s); (i) > (e); (i)--) #define debug(x) cout << #x << ": " << x << endl #define mp make_pair #define pb push_back const ll MOD = 1000000007; const int INF = 1e9; const ll LINF = 1e16; const double PI = acos(-1.0); int dx[8] = {0, 0, 1, -1, 1, 1, -1, -1}; int dy[8] = {1, -1, 0, 0, 1, -1, 1, -1}; /* ----- xtimex Problem: / Link: ----- */ /* ------問題------ -----問題ここまで----- */ /* -----解説等----- ----解説ここまで---- */ ll N,target; ll a[50]; bool memo[51][100110]; ll ans = 0LL; string s=""; void f(int i,int sum) { //cout << s << endl; if (i == N&&sum == target) { cout << s << endl; exit(0); } if (i >= N || sum >= target)return; if (memo[i][sum])return ; memo[i][sum] = 1; s+= '+'; f(i + 1, sum + a[i]); s.erase(s.size() - 1); s += '*'; f(i + 1, sum * a[i]); s.erase(s.size() - 1); return; } int main() { cin.tie(0); ios_base::sync_with_stdio(false); FOR(i,0,50)FOR(j,0,100011) memo[i][j]=0; cin >> N>>target; FOR(i, 0, N)cin >> a[i]; f(1, a[0]); return 0; }