A triangular number is the number of dots in an equilateral triangle uniformly filled with dots. For example, three dots can be arranged in a triangle; thus three is a triangular number. The n-th triangular number is the number of dots in a triangle with n dots on a side. <image>. You can learn more about these numbers from Wikipedia (http://en.wikipedia.org/wiki/Triangular_number).

Your task is to find out if a given integer is a triangular number.

Input

The first line contains the single number n (1 ≤ n ≤ 500) — the given integer.

Output

If the given integer is a triangular number output YES, otherwise output NO.

Examples

Input

1


Output

YES


Input

2


Output

NO


Input

3


Output

YES
#include <bits/stdc++.h>
using namespace std;
void bug() { cout << "here\n"; }
template <typename T>
inline T S(T a) {
  return a * a;
}
template <class T>
inline string tostring(T a) {
  ostringstream os("");
  os << a;
  return os.str();
}
template <typename T>
inline long long tolong(T a) {
  long long res;
  istringstream os(a);
  os >> res;
  return res;
}
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <typename T>
inline T bigmod(T a, T b, T m) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return S(bigmod(a, b / 2, m)) % m;
  else
    return (a % m * bigmod(a, b - 1, m)) % m;
}
const int inf = (int)1e9 + 5;
const long long linf = (long long)1e16 + 5;
const long long modd = (long long)1e9 + 7;
const int mod = 10000007;
void pr(int f) {
  if (f == 1)
    cout << "YES\n";
  else
    cout << "NO\n";
}
bool isprime(long long n) {
  if (n < 2) return 0;
  if (n == 2)
    return 1;
  else {
    for (int i = 2; i * i <= n; i++)
      if (n % i == 0) return 0;
  }
  return 1;
}
long long countSum(long long x) {
  long long res = 0;
  while (x != 0) {
    res += (x % 10);
    x /= 10;
  }
  return res;
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    long long res = power(a, b / 2);
    if (b % 2)
      return res * a * res;
    else
      return res * res;
  }
}
int cnt[3];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  int x = sqrt(n * 2);
  cout << (x * (x + 1) / 2 == n ? "YES" : "NO") << endl;
}
