#include <algorithm>
#include <cmath>
#include <iostream>
#include <vector>

#define REP(i,s,n) for(int i=(int)(s);i<(int)(n);i++)

using namespace std;
typedef vector<int> VI;


/**
 * Segment Tree. This data structure is useful for fast folding on intervals of an array
 * whose elements are elements of monoid M. Note that constructing this tree requires the identity
 * element of M and the operation of M.
 * Header requirement: vector
 */
template<class I, class BiOp = I (*) (I, I)>
class SegTree {
  int n;
  std::vector<I> dat;
  BiOp op;
  I e;
public:
  SegTree(int n_, BiOp op, I e) : op(op), e(e) {
    n = 1;
    while (n < n_) { n *= 2; } // n is a power of 2
    dat.resize(2 * n);
    for (int i = 0; i < 2 * n - 1; i++) {
      dat[i] = e;
    }
  }
  /* ary[k] <- v */
  void update(int k, I v) {
    k += n - 1;
    dat[k] = v;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = op(dat[2 * k + 1], dat[2 * k + 2]);
    }
  }
  void update_array(int k, int len, const I *vals) {
    for (int i = 0; i < len; ++i) {
      update(k + i, vals[i]);
    }
  }
  /*
    Updates all elements. O(n)
   */
  void update_all(const I *vals, int len) {
    for (int k = 0; k < min(n, len); ++k) {
      dat[k + n - 1] = vals[k];
    }
    for (int k = min(n, len); k < n; ++k) {
      dat[k + n - 1] = e;
    }
    for (int b = n / 2; b >= 1; b /= 2) {
      for (int k = 0; k < b; ++k) {
	dat[k + b - 1] = op(dat[k * 2 + b * 2 - 1], dat[k * 2 + b * 2]);
      }
    }
  }
  /* l,r are for simplicity */
  I querySub(int a, int b, int k, int l, int r) const {
    // [a,b) and  [l,r) intersects?
    if (r <= a || b <= l) return e;
    if (a <= l && r <= b) return dat[k];
    I vl = querySub(a, b, 2 * k + 1, l, (l + r) / 2);
    I vr = querySub(a, b, 2 * k + 2, (l + r) / 2, r);
    return op(vl, vr);
  }
  /* [a, b] (note: inclusive) */
  I query(int a, int b) const {
    return querySub(a, b + 1, 0, 0, n);
  }
};

const int N = 5000001;
int tbl[N];

int dset(int v) {
  int bits = 0;
  while (v > 0) {
    bits |= 1 << (v % 10);
    v /= 10;
  }
  return bits;
}

struct bor {
  int operator()(int a, int b) const { return a | b; }
};

void solve(int bits) {
  tbl[0] = tbl[1] = 0;
  
  REP(i, 2, N) {
    tbl[i] = dset(i);
  }
  REP(i, 2, sqrt(N) + 1) {
    if (tbl[i]) {
      REP(j, 2, (N + i - 1) / i) {
	tbl[i * j] = 0;
      }
    }
  }
  SegTree<int, bor> st(N, bor(), 0);
  st.update_all(tbl, N);
  int ma = -1;
  REP(i, 1, N) {
    int acc = 0;
    int j = i;
    for (; j < N; ++j) {
      if (tbl[j] & ~bits) {
	break;
      }
      acc |= tbl[j];
      if (acc == bits) {
	ma = max(ma, j - i);
      }
    }
    i = j;
  }
  cout << ma << endl;
}

int main(void){
  int sn;
  cin >> sn;
  int bits = 0;
  REP(i, 0, sn) {
    int y;
    cin >> y;
    bits |= 1 << y;
  }
  solve(bits);
}