Problem: Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.

More formally, you are given an odd numer n. Find a set of numbers pi (1 ≤ i ≤ k), such that

  1. 1 ≤ k ≤ 3
  2. pi is a prime
  3. <image>



The numbers pi do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.

Input

The single line contains an odd number n (3 ≤ n < 109).

Output

In the first line print k (1 ≤ k ≤ 3), showing how many numbers are in the representation you found.

In the second line print numbers pi in any order. If there are multiple possible solutions, you can print any of them.

Examples

Input

27


Output

3
5 11 11

Note

A prime is an integer strictly larger than one that is divisible only by one and by itself.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int M = 2000;
const int inf = 0x3f3f3f3f;
const double eps = 1e-9;
int MOD = 1e9 + 7;
int cas = 1;
int n, m, k, q, L;
int sgn(double x) { return x < -eps ? -1 : x < eps ? 0 : 1; }
struct Point {
  double x, y;
  double v, l;
  Point() {}
  Point(double _x, double _y) {
    x = _x;
    y = _y;
  }
};
double cross(Point a, Point b, Point c) {
  return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}
bool same(Point a, Point b) {
  if (sgn(a.x - b.x) == 0 && sgn(a.y - b.y) == 0) return true;
  return false;
}
double dist(Point a, Point b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
bool cmp(const Point &a, const Point &b) {
  if (sgn(a.y - b.y) == 0) return a.x < b.x;
  return a.y < b.y;
}
Point tmp;
bool cmp2(const Point &a, const Point &b) {
  int d = sgn(cross(tmp, a, b));
  if (!d)
    return dist(tmp, a) < dist(tmp, b);
  else {
    return d > 0;
  }
}
Point p[20], convex[20];
double solve(Point p[], int n) {
  if (n == 1) return 0;
  if (n == 2) return dist(p[0], p[1]);
  int mi = 0;
  for (int i = 1; i < n; i++)
    if (cmp(p[i], p[mi])) mi = i;
  swap(p[mi], p[0]);
  tmp = p[0];
  sort(p + 1, p + n, cmp2);
  int top = 0;
  convex[top++] = p[0];
  convex[top++] = p[1];
  int k = 2;
  while (k <= n) {
    while (top >= 2 &&
           sgn(cross(convex[top - 2], convex[top - 1], p[k % n])) >= 0)
      top--;
    convex[top++] = p[k % n];
    k++;
  }
  double ans = 0;
  if (top <= 1)
    return 0;
  else if (top == 2)
    return dist(convex[0], convex[1]);
  else {
    for (int i = 0; i < top; i++) ans += dist(convex[i], convex[(i + 1) % top]);
  }
  return top;
}
Point tp[20];
long long qmod(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
bool pr(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
int main() {
  int T;
  int n, t;
  cin >> n;
  if (pr(n))
    printf("1\n%d", n);
  else if (pr(n - 2))
    printf("2\n2 %d", n - 2);
  else if (pr(n - 4))
    printf("3\n2 2 %d", n - 4);
  else {
    int i;
    n -= 3;
    for (i = 3;; i += 2)
      if (pr(i) && pr(n - i)) break;
    printf("3\n3 %d %d", i, n - i);
  }
  return 0;
}