#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i)) #define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i)) #define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i)) #if defined(_MSC_VER) || __cplusplus > 199711L #define aut(r,v) auto r = (v) #else #define aut(r,v) __typeof(v) r = (v) #endif #define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it) #define all(o) (o).begin(), (o).end() #define pb(x) push_back(x) #define mp(x,y) make_pair((x),(y)) #define mset(m,v) memset(m,v,sizeof(m)) #define INF 0x3f3f3f3f #define INFL 0x3f3f3f3f3f3f3f3fLL using namespace std; typedef vector vi; typedef pair pii; typedef vector > vpii; typedef long long ll; template inline void amin(T &x, U y) { if(y < x) x = y; } template inline void amax(T &x, U y) { if(x < y) x = y; } vector primes; vector smallestPrimeFactor; void linearSieve(int n) { if(n < 1) n = 1; if((int)smallestPrimeFactor.size() >= n+1) return; int primePiBound = n < 20 ? n - 1 : (int)(n / (log(n * 1.) - 2) + 2); primes.assign(primePiBound + 1, numeric_limits::max()); int P = 0; smallestPrimeFactor.assign(n + 1, 0); smallestPrimeFactor[1] = 1; int n2 = n / 2, n3 = n / 3, n5 = n / 5; if(n >= 2) primes[P ++] = 2; if(n >= 3) primes[P ++] = 3; for(int q = 2; q <= n; q += 2) smallestPrimeFactor[q] = 2; for(int q = 3; q <= n; q += 6) smallestPrimeFactor[q] = 3; for(int q = 5; q <= n5; q += 2) { if(smallestPrimeFactor[q] == 0) primes[P ++] = smallestPrimeFactor[q] = q; int bound = smallestPrimeFactor[q]; for(int i = 2; ; ++ i) { int p = primes[i]; if(p > bound) break; int pq = p * q; if(pq > n) break; smallestPrimeFactor[pq] = p; } } for(int q = (n5 + 1) | 1; q <= n; q += 2) { if(smallestPrimeFactor[q] == 0) primes[P ++] = smallestPrimeFactor[q] = q; } primes.resize(P); } int getdigits(int x) { int r = 0; while(x > 0) r |= 1 << (x % 10), x /= 10; return r; } int main() { const int X = 5000000; linearSieve(X + 1000); int N; while(cin >> N) { int des = 0; rep(i, N) { int A; cin >> A; des |= 1 << A; } int ans = -1; int cnt[10] = {}; for(int i = 0, j = 0; primes[i] <= X; ++ i) { for(; primes[j] < X; ++ j) { int t = getdigits(primes[j]); if(t & ~des) break; rep(d, 10) if(t >> d & 1) ++ cnt[d]; } if(i < j && 10 - count(cnt, cnt + 10, 0) == N) amax(ans, min(X, primes[j] - 1) - (i == 0 ? 1 : primes[i-1] + 1)); if(i == j) { ++ j; }else { int t = getdigits(primes[i]); rep(d, 10) if(t >> d & 1) -- cnt[d]; } } cout << ans << endl; } return 0; }