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;
int ri() {
  int r;
  cin >> r;
  return r;
}
unsigned int rui() {
  unsigned int r;
  cin >> r;
  return r;
}
long long rl() {
  long long r;
  cin >> r;
  return r;
}
unsigned long long rul() {
  unsigned long long r;
  cin >> r;
  return r;
}
double rd() {
  double r;
  cin >> r;
  return r;
}
string rss() {
  string r;
  cin >> r;
  return r;
}
string rs() {
  string r;
  getline(cin, r);
  return r;
}
vector<string> rvs(int n) {
  assert(n);
  vector<string> r;
  for (int i = 0; i < n; ++i) {
    string s = rs();
    while (s.empty()) s = rs();
    r.push_back(s);
  }
  return r;
}
vector<int> rvi(int n) {
  assert(n);
  vector<int> r;
  for (int i = 0; i < n; ++i) r.push_back(ri());
  return r;
}
vector<unsigned int> rvui(int n) {
  assert(n);
  vector<unsigned int> r;
  for (int i = 0; i < n; ++i) r.push_back(rui());
  return r;
}
vector<long long> rvl(int n) {
  assert(n);
  vector<long long> r;
  for (int i = 0; i < n; ++i) r.push_back(rl());
  return r;
}
vector<unsigned long long> rvul(int n) {
  assert(n);
  vector<unsigned long long> r;
  for (int i = 0; i < n; ++i) r.push_back(rul());
  return r;
}
vector<double> rvd(int n) {
  assert(n);
  vector<double> r;
  for (int i = 0; i < n; ++i) r.push_back(rd());
  return r;
}
int main() {
  int v = ri();
  for (int i = 1; i < 1000; ++i) {
    auto cur = i * (i + 1) / 2;
    if (cur == v) {
      cout << "YES" << endl;
      return 0;
    } else if (cur > v)
      break;
  }
  cout << "NO" << endl;
  return 0;
}
