#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
inline void relax_min(T &a, U b) {
  a = min(a, b);
}
template <class T, class U>
inline void relax_max(T &a, U b) {
  a = max(a, b);
}
template <typename T>
inline bool updmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool updmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &a) {
  in >> a.first >> a.second;
  return in;
}
template <class T, class U>
ostream &operator<<(ostream &out, const pair<T, U> &a) {
  out << a.first << ' ' << a.second;
  return out;
}
template <class T>
istream &operator>>(istream &in, vector<T> &a) {
  for (decltype(0)((i)) = (0); ((i)) < ((((int)(a).size()))); ++((i)))
    in >> a[i];
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &a) {
  for (decltype(0)((i)) = (0); ((i)) < ((((int)(a).size()))); ++((i)))
    out << a[i] << ' ';
  return out;
}
int next_int() {
  int(tmp);
  scanf("%d", &(tmp));
  return tmp;
}
long long next_ll() {
  long long(tmp);
  scanf("%lld", &(tmp));
  return tmp;
}
long long modPow(long long a, long long b, long long MOD) {
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1)
      ret = ((((((long long)(ret) * (long long)(a))) % ((MOD))) + ((MOD))) %
             ((MOD)));
    a = ((((((long long)(a) * (long long)(a))) % ((MOD))) + ((MOD))) % ((MOD)));
  }
  return ret;
}
const long long EXACTLY_ONE_HUNDRED = 100;
const long long EXACTLY_ONE_THOUSAND = 1000;
const long long EXACTLY_ONE_MILLION = 1000 * 1000;
const long long EXACTLY_ONE_BILLION =
    EXACTLY_ONE_THOUSAND * EXACTLY_ONE_MILLION;
const int MAXN1 = (int)(EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN2 = (int)(2 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN3 = (int)(3 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN4 = (int)(4 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN5 = (int)(5 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN = MAXN3;
const int INF = (int)EXACTLY_ONE_BILLION + 7;
const int INF2 = 1 << 30;
const long long INFLL = 4LL * EXACTLY_ONE_BILLION * EXACTLY_ONE_BILLION + 7;
const long long INFLL2 = 1LL << 60;
const long double Pi = acos(-1.0);
const long double EPS = 1e-7;
const int MOD = (int)EXACTLY_ONE_BILLION + 7;
int main() {
  int(n);
  scanf("%d", &(n));
  vector<int> l(n), r(n);
  for (decltype(0)((IT_FOR_SCANNING)) = (0); ((IT_FOR_SCANNING)) < ((((n))));
       ++((IT_FOR_SCANNING)))
    scanf("%d%d", &(((l))[IT_FOR_SCANNING]), &(((r))[IT_FOR_SCANNING]));
  vector<int> p(n);
  for (decltype(0)((i)) = (0); ((i)) < ((n)); ++((i))) {
    p[i] = i;
  }
  auto cmp = [&l, &r](int i, int j) { return l[i] < l[j]; };
  sort((p).begin(), (p).end(), cmp);
  int last1 = -1, last2 = -1;
  for (decltype(0)((I)) = (0); ((I)) < ((n)); ++((I))) {
    int i = p[I];
    if (l[i] > last1) {
      last1 = r[i];
    } else if (l[i] > last2) {
      last2 = r[i];
    } else {
      puts("NO");
      return 0;
    }
  }
  puts("YES");
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

Polycarp is a great fan of television.

He wrote down all the TV programs he is interested in for today. His list contains n shows, i-th of them starts at moment li and ends at moment ri.

Polycarp owns two TVs. He can watch two different shows simultaneously with two TVs but he can only watch one show at any given moment on a single TV. If one show ends at the same moment some other show starts then you can't watch them on a single TV.

Polycarp wants to check out all n shows. Are two TVs enough to do so?

Input

The first line contains one integer n (1 ≤ n ≤ 2·105) — the number of shows.

Each of the next n lines contains two integers li and ri (0 ≤ li < ri ≤ 109) — starting and ending time of i-th show.

Output

If Polycarp is able to check out all the shows using only two TVs then print "YES" (without quotes). Otherwise, print "NO" (without quotes).

Examples

Input

3
1 2
2 3
4 5


Output

YES


Input

4
1 2
2 3
2 3
1 2


Output

NO

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
inline void relax_min(T &a, U b) {
  a = min(a, b);
}
template <class T, class U>
inline void relax_max(T &a, U b) {
  a = max(a, b);
}
template <typename T>
inline bool updmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool updmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &a) {
  in >> a.first >> a.second;
  return in;
}
template <class T, class U>
ostream &operator<<(ostream &out, const pair<T, U> &a) {
  out << a.first << ' ' << a.second;
  return out;
}
template <class T>
istream &operator>>(istream &in, vector<T> &a) {
  for (decltype(0)((i)) = (0); ((i)) < ((((int)(a).size()))); ++((i)))
    in >> a[i];
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &a) {
  for (decltype(0)((i)) = (0); ((i)) < ((((int)(a).size()))); ++((i)))
    out << a[i] << ' ';
  return out;
}
int next_int() {
  int(tmp);
  scanf("%d", &(tmp));
  return tmp;
}
long long next_ll() {
  long long(tmp);
  scanf("%lld", &(tmp));
  return tmp;
}
long long modPow(long long a, long long b, long long MOD) {
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1)
      ret = ((((((long long)(ret) * (long long)(a))) % ((MOD))) + ((MOD))) %
             ((MOD)));
    a = ((((((long long)(a) * (long long)(a))) % ((MOD))) + ((MOD))) % ((MOD)));
  }
  return ret;
}
const long long EXACTLY_ONE_HUNDRED = 100;
const long long EXACTLY_ONE_THOUSAND = 1000;
const long long EXACTLY_ONE_MILLION = 1000 * 1000;
const long long EXACTLY_ONE_BILLION =
    EXACTLY_ONE_THOUSAND * EXACTLY_ONE_MILLION;
const int MAXN1 = (int)(EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN2 = (int)(2 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN3 = (int)(3 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN4 = (int)(4 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN5 = (int)(5 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN = MAXN3;
const int INF = (int)EXACTLY_ONE_BILLION + 7;
const int INF2 = 1 << 30;
const long long INFLL = 4LL * EXACTLY_ONE_BILLION * EXACTLY_ONE_BILLION + 7;
const long long INFLL2 = 1LL << 60;
const long double Pi = acos(-1.0);
const long double EPS = 1e-7;
const int MOD = (int)EXACTLY_ONE_BILLION + 7;
int main() {
  int(n);
  scanf("%d", &(n));
  vector<int> l(n), r(n);
  for (decltype(0)((IT_FOR_SCANNING)) = (0); ((IT_FOR_SCANNING)) < ((((n))));
       ++((IT_FOR_SCANNING)))
    scanf("%d%d", &(((l))[IT_FOR_SCANNING]), &(((r))[IT_FOR_SCANNING]));
  vector<int> p(n);
  for (decltype(0)((i)) = (0); ((i)) < ((n)); ++((i))) {
    p[i] = i;
  }
  auto cmp = [&l, &r](int i, int j) { return l[i] < l[j]; };
  sort((p).begin(), (p).end(), cmp);
  int last1 = -1, last2 = -1;
  for (decltype(0)((I)) = (0); ((I)) < ((n)); ++((I))) {
    int i = p[I];
    if (l[i] > last1) {
      last1 = r[i];
    } else if (l[i] > last2) {
      last2 = r[i];
    } else {
      puts("NO");
      return 0;
    }
  }
  puts("YES");
  return 0;
}

```