import math
import sys
sys.setrecursionlimit(10**7)
#競技プログラミング対整数問題のライブラリーです
class integerlib():
    def __init__(self):
        pass
    
    def primeset(self,N): #N以下の素数をsetで求める.エラトステネスの篩O(√Nlog(N))
        lsx = [1]*(N+1)
        for i in range(2,int(-(-N**0.5//1))+1):
            if lsx[i] == 1:
                for j in range(i,N//i+1):
                    lsx[j*i] = 0
        setprime = set()
        for i in range(2,N+1):
            if lsx[i] == 1:
                setprime.add(i)
        return setprime
    
    def defprime(self,N):#素数かどうかの判定、エラトステネスの篩O(√Nlog(N))
        return N in self.primeset(N)
    
    def gcd(self,ls):#最大公約数
        ls = list(ls)
        ans = 0
        for i in ls:
            ans = math.gcd(ans,i)
        return ans

    def lmc(self,ls):#最小公倍数
        ls = list(ls)
        ans = self.gcd(ls)
        for i in ls:
            ans = self.lmcsub(ans,i)
        return ans
    
    def lmcsub(self,a,b):
        gcd = math.gcd(a,b)
        lmc = (a*b)//gcd
        return lmc

    
    def factorization(self,N):#素因数分解√N
        arr = []
        temp = N
        for i in range(2, int(-(-N**0.5//1))+1):
            if temp%i==0:
                cnt=0
                while temp%i==0:
                    cnt+=1
                    temp //= i
                arr.append([i, cnt])
        if temp!=1:
            arr.append([temp, 1])
        if arr==[]:
            arr.append([N, 1])
        return arr #[素因数、個数]

    def factorizationset(self,N):#素因数分解√N,含まれている素因数の種類
        if N == 1:
            return set()
        ls = self.factorization(N)
        setf = set()
        for j in ls:
            setf.add(j[0])
        return setf

    def divisorsnum(self,N):#約数の個数
        ls = []
        for i in self.factorization(N):
            ls.append(i[1])
        d = 1
        for i in ls:
            d *= i+1
        return d

    def Eulerfunc(self,N):#オイラー関数正の整数Nが与えられる。1,2,…,Nのうち、Nと互いに素であるものの個数を求めよ。
        ls = list(self.factorizationset(N))
        ls2 = [N]
        for i in ls:
            ls2.append(ls2[-1]-ls2[-1]//i)
        return ls2[-1]

    def make_divisors(self,N):#約数列挙O(√N)
        lower_divisors , upper_divisors = [], []
        i = 1
        while i*i <= N:
            if N % i == 0:
                lower_divisors.append(i)
                if i != N // i:
                    upper_divisors.append(N//i)
            i += 1
        return lower_divisors + upper_divisors[::-1]

    def invmod(self,a,mod):#mod逆元
        if a == 0:
            return 0
        if a == 1:
            return 1
        return (-self.invmod(mod % a, mod) * (mod // a)) % mod
        
    def cmbmod(self,n, r, mod):#nCr % mod
        inv = [0,1]
        for i in range(2, n + 1):
            inv.append((-inv[mod % i] * (mod // i)) % mod)
        cmd = 1
        for i in range(1,min(r,n-r)+1):
            cmd = (cmd*(n-i+1)*inv[i])%mod
        return cmd

    def permmod(self,n, r, mod):#nPr % mod
        perm = 1
        for i in range(n,r-1,-1):
            perm = (perm*i)%mod
        return perm

    def modPow(self,a,n,mod):#繰り返し二乗法 a**n % mod
        if n==0:
            return 1
        if n==1:
            return a%mod
        if n % 2 == 1:
            return (a*self.modPow(a,n-1,mod)) % mod
        t = self.modPow(a,n//2,mod)
        return (t*t)%mod
import copy
#5000000までの素数列挙
#10種類の数字をカウント
IT = integerlib()
lsp = list(IT.primeset(5000000))
lsp.sort()
M = len(lsp)
#尺取り法
N = int(input())
lsA = set(list(map(int,input().split())))
counter = [0]*(10)
def ok(counter):#余計なのはいってたらf
    f = True
    for i in range(10):
        if counter[i] > 0 and (not(i in lsA)):
            f = False
    return f
def ok2(counter):#条件を満たす
    f = True
    for i in range(10):
        if counter[i] > 0 and (not(i in lsA)):
            f = False
        if counter[i] == 0 and (i in lsA):
            f = False
    return f
r = 0
ans = -1
for i in range(M):
    while r < M and ok(counter):
        c = list(str(lsp[r]))
        c = [int(j) for j in c]
        for j in c:
            counter[j] += 1
        if ok2(counter):
            if i == 0:
                ln = 1
            else:
                ln = (lsp[i-1]+1)
            if r == M-1:
                rn = 5000000
            else:
                rn = lsp[r+1]-1
            ans = max(ans,rn-ln)        
        r += 1
    if ok2(counter):
        if i == 0:
            ln = 1
        else:
            ln = (lsp[i-1]+1)
        if r == len(lsp):
            rn = 5000000
        else:
            rn = lsp[r+1]-1
        ans = max(ans,rn-ln)
    c = list(str(lsp[i]))
    c = [int(j) for j in c]
    for j in c:
        counter[j] -= 1
print(ans)