/*
  Author : MatsuTaku
  Date   : 03/25/22
*/

#include <bits/stdc++.h>
#include <x86intrin.h>
//#include <atcoder/all>
#define _overload3(_1, _2, _3, target, ...) target
#define _REP(i, l, r)   for (int i = (l), i##_less = (r); i < i##_less; i++)
#define _rep(i, n)      _REP(i, 0, n)
#define rep(args...)    _overload3(args, _REP, _rep)(args)
#define _RREP(i, l, r)  for (int i = (r)-1, i##_least = (l); i >= i##_least; i--)
#define _rrep(i, n)     _RREP(i, 0, n)
#define rrep(args...)   _overload3(args, _RREP, _rrep)(args)
#define chmax(dst, x)   dst = max(dst, (x))
#define chmin(dst, x)   dst = min(dst, (x))
using namespace std;
using lint = long long int;
using ulint = unsigned long long int;
template<typename T>
using vvec = vector<vector<T>>;
template<typename T>
vvec<T> make_vvec(int n, int m, T v) { return vvec<T>(n, vector<T>(m, v)); }
template<typename T>
using min_queue = priority_queue<T, vector<T>, greater<T>>;

class Solver {
 public:
  Solver();
  void solve();
};

Solver::Solver() {}

void Solver::solve() {
  int n; cin>>n;
  vector<lint> A(n);
  rep(i, n) cin>>A[i];
  sort(A.begin(), A.end());
  int k = 0;
  while (k < n and A[k] == 0)
    k++;
  rrep(i, k, n) A[i] -= A[k];
  lint g = A[k+1];
  rep(i, k+1, n) {
    g = gcd(g, A[i]);
  }
  bool ans;
  if (k < n-1 and A[k]==A[k+1]) {
    ans = A[n-1] == A[k];
  } else {
    lint c = 0;
    rep(i, k, n-1) {
      lint d = A[i+1]-A[i];
      c += d/g-1;
    }
    ans = c <= k;
  }
  cout << (ans ? "Yes" : "No") << endl;
}

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(10);
  Solver solver;
  int t = 1;
//  cin>>t;
  while (t--) {
    solver.solve();
  }
  return 0;
}