Phoenix is trying to take a photo of his n friends with labels 1, 2, ..., n who are lined up in a row in a special order. But before he can take the photo, his friends get distracted by a duck and mess up their order.

Now, Phoenix must restore the order but he doesn't remember completely! He only remembers that the i-th friend from the left had a label between a_i and b_i inclusive. Does there exist a unique way to order his friends based of his memory? 

Input

The first line contains one integer n (1 ≤ n ≤ 2⋅10^5) — the number of friends.

The i-th of the next n lines contain two integers a_i and b_i (1 ≤ a_i ≤ b_i ≤ n) — Phoenix's memory of the i-th position from the left.

It is guaranteed that Phoenix's memory is valid so there is at least one valid ordering.

Output

If Phoenix can reorder his friends in a unique order, print YES followed by n integers — the i-th integer should be the label of the i-th friend from the left.

Otherwise, print NO. Then, print any two distinct valid orderings on the following two lines. If are multiple solutions, print any.

Examples

Input


4
4 4
1 3
2 4
3 4


Output


YES
4 1 2 3 


Input


4
1 3
2 4
3 4
2 3


Output


NO
1 3 4 2 
1 2 4 3 
4
3 4
1 2
3 4
1 2
 4
2 4
1 3
4 4
1 1
 10
6 6
1 2
2 2
6 7
8 8
3 5
5 5
4 4
8 9
10 10
 4
1 4
1 4
2 4
1 2
 8
3 4
7 8
2 6
1 3
6 7
5 8
4 6
8 8
 7
7 7
2 3
2 3
5 7
5 6
4 4
1 2
 15
2 2
10 11
2 5
6 7
7 12
9 9
2 3
5 14
1 3
10 10
15 15
12 12
5 6
6 6
14 15
 5
1 3
4 4
5 5
1 4
1 5
 4
1 4
2 3
2 3
4 4
 4
1 4
1 4
1 2
1 4
 5
1 3
1 3
1 5
3 4
1 5
 3
1 3
2 2
1 3
 5
4 4
1 4
3 4
1 3
3 5
 3
1 3
1 2
2 3
 4
2 4
1 3
2 3
1 2
 5
4 5
4 5
1 3
1 2
2 4
 6
2 6
4 6
6 6
2 2
3 3
1 3
 3
2 3
1 2
1 3
 6
4 6
1 3
4 6
4 4
1 4
1 2
 7
3 6
6 6
5 7
2 4
2 3
1 2
3 5
 8
4 6
8 8
1 4
4 5
6 8
1 2
3 6
1 4
 6
2 4
2 6
4 5
2 5
6 6
1 4
 8
4 4
7 7
3 5
5 7
1 8
4 5
1 5
2 2
 4
3 4
1 3
2 3
1 2
 10
9 9
3 4
1 1
5 5
7 7
3 7
6 10
3 3
10 10
1 3
 4
1 4
4 4
2 3
1 1
 1
1 1
 8
1 5
6 8
1 3
4 8
1 1
3 3
6 6
8 8
 5
3 3
4 4
2 2
1 1
5 5
 4
1 1
2 3
2 3
1 4
 4
3 4
1 4
2 3
1 2
 2
1 2
1 2
 8
3 3
3 5
2 3
1 1
5 7
7 7
5 5
3 8
 5
2 2
3 3
1 2
1 5
4 4
 6
1 3
2 4
3 4
2 3
5 5
6 6
 6
1 4
3 6
6 6
3 5
1 5
2 5
 5
4 4
2 3
4 5
1 4
2 4
 3
1 2
1 2
1 3
 4
1 4
2 2
3 3
1 4
 5
2 5
1 1
2 3
3 4
5 5
 4
2 4
1 4
3 4
1 3
 6
1 1
4 6
3 6
1 3
3 4
3 6
 5
2 3
3 4
4 5
1 1
2 3
 5
2 3
1 5
2 4
3 5
4 4
 20
12 14
13 13
18 18
3 3
7 8
11 11
10 13
2 2
9 10
3 5
7 8
15 17
14 15
9 9
20 20
15 16
6 6
19 19
1 1
5 5
 5
1 5
2 2
3 3
4 4
1 5
 7
5 7
1 5
1 2
1 5
5 6
3 5
7 7

NO
3 1 4 2
3 2 4 1
 NO
3 2 4 1 
2 3 4 1  YES
6 1 2 7 8 3 5 4 9 10  NO
2 3 4 1 
1 3 4 2 
 YES
3 7 2 1 6 5 4 8  NO
7 2 3 6 5 4 1 
7 3 2 6 5 4 1 
 YES
2 11 4 7 8 9 3 13 1 10 15 12 5 6 14  NO
1 4 5 2 3 
2 4 5 1 3  NO
1 2 3 4 
1 3 2 4 
 NO
2 3 1 4 
1 3 2 4 
 NO
1 2 4 3 5 
2 1 4 3 5 
 NO
1 2 3 
3 2 1 
 NO
4 2 3 1 5 
4 1 3 2 5  NO
2 1 3
1 2 3
 NO
4 2 3 1
4 1 3 2
 NO
4 5 2 1 3
4 5 1 2 3
 NO
4 5 6 2 3 1 
5 4 6 2 3 1 
 NO
2 1 3 
3 1 2 
 NO
5 2 6 4 3 1 
5 1 6 4 3 2 
 NO
5 6 7 3 2 1 4 
5 6 7 2 3 1 4  NO
5 8 2 4 7 1 6 3
5 8 1 4 7 2 6 3
 NO
2 5 4 3 6 1 
3 5 4 2 6 1  YES
4 7 3 6 8 5 1 2  NO
4 2 3 1
4 1 3 2
 YES
9 4 1 5 7 6 8 3 10 2  NO
3 4 2 1 
2 4 3 1  YES
1  NO
4 7 2 5 1 3 6 8 
5 7 2 4 1 3 6 8  YES
3 4 2 1 5  NO
1 2 3 4 
1 3 2 4 
 NO
3 4 2 1 
4 3 2 1 
 NO
1 2 
2 1 
 YES
3 4 2 1 6 7 5 8  YES
2 3 1 5 4  NO
1 3 4 2 5 6
1 2 4 3 5 6
 NO
1 5 6 3 2 4 
2 5 6 3 1 4 
 NO
4 2 5 1 3 
4 3 5 1 2  NO
1 2 3 
2 1 3 
 NO
1 2 3 4 
4 2 3 1 
 NO
4 1 2 3 5 
3 1 2 4 5 
 NO
2 3 4 1
2 1 4 3
 NO
1 4 5 2 3 6 
1 5 4 2 3 6 
 NO
2 4 5 1 3 
3 4 5 1 2 
 NO
2 1 3 5 4 
3 1 2 5 4  NO
14 13 18 3 7 11 12 2 10 4 8 17 15 9 20 16 6 19 1 5 
14 13 18 3 8 11 12 2 10 4 7 17 15 9 20 16 6 19 1 5 
 NO
1 2 3 4 5 
5 2 3 4 1 
 NO
6 2 1 3 5 4 7
6 1 2 3 5 4 7

#include <bits/stdc++.h>
using namespace std;
const int NMAX = 2e5 + 100;
int n, m, k;
vector<int> g[2 * NMAX];
int used[2 * NMAX];
int l[2 * NMAX];
void bfs() {
  queue<int> q;
  for (int i = 0; i < m + n; i++) l[i] = -1;
  for (int i = 0; i < n; i++)
    if (used[i] == -1) q.push(i), l[i] = 0;
  k = 1e9;
  while (!q.empty()) {
    int v = q.front();
    if (l[v] >= k) return;
    if (l[v] % 2 == 0) {
      for (int i : g[v])
        if (l[i] == -1) {
          q.push(i);
          l[i] = l[v] + 1;
          if (used[i] == -1) k = l[i];
        }
    } else if (l[used[v]] == -1) {
      if (used[v] == -1) cout << "WARNING\n";
      l[used[v]] = l[v] + 1;
      q.push(used[v]);
    }
    q.pop();
  }
}
bool dfs(int v) {
  if (v == -1) return true;
  for (int to : g[v])
    if ((used[to] == -1) or (l[used[to]] == l[v] - 2))
      if (dfs(used[to])) {
        used[to] = v;
        used[v] = to;
        return true;
      }
  l[v] = -1;
  return false;
}
void SARA() {
  for (int i = 0; i < n + m; i++) used[i] = -1;
  while (true) {
    bfs();
    if (k == 1e9) break;
    for (int i = 0; i < m + n; i++)
      if ((l[i] == k) and (used[i] == -1)) dfs(i);
  }
}
void add_edge(int i, int j) {
  g[i - 1].push_back(n + j - 1);
  g[n + j - 1].push_back(i - 1);
}
int N;
int lt[NMAX];
int rt[NMAX];
set<vector<int> > answers;
void add(vector<int> ans) {
  answers.insert(ans);
  if (answers.size() > 1) {
    cout << "NO\n";
    for (auto it = answers.begin(); it != answers.end(); it++, cout << "\n")
      for (int i : *it) cout << i << " ";
    exit(0);
  }
}
void try_() {
  for (int i = 0; i < 2 * N; i++) g[i].clear();
  vector<int> pr;
  for (int i = 1; i <= n; i++) pr.push_back(i);
  random_shuffle(pr.begin(), pr.end());
  vector<int> bck(n);
  for (int i = 0; i < n; i++) bck[pr[i] - 1] = i + 1;
  for (int i = 1; i <= N; i++) {
    if (rt[i] - lt[i] < 6)
      for (int j = lt[i]; j <= rt[i]; j++) add_edge(pr[i - 1], pr[j - 1]);
    else {
      set<int> st;
      while (st.size() < 5) st.insert(rand() % (rt[i] - lt[i] + 1) + lt[i]);
      for (int j : st) add_edge(pr[i - 1], pr[j - 1]);
    }
  }
  SARA();
  int q = 0;
  for (int i = 0; i < n; i++)
    if (used[i] != -1) q++;
  if (q < n) return;
  vector<int> ans(n);
  for (int i = 0; i < n; i++) ans[i] = bck[used[pr[i] - 1] - n];
  add(ans);
}
void gridi() {
  vector<int> ans(n);
  set<pair<pair<int, int>, int> > st;
  set<pair<int, int> > bef;
  for (int i = 1; i <= n; i++) st.insert({{lt[i], rt[i]}, i - 1});
  for (int i = 1; i <= n; i++) {
    while ((st.size() > 0) and ((*st.begin()).first.first <= i)) {
      bef.insert({(*st.begin()).first.second, (*st.begin()).second});
      st.erase(st.begin());
    }
    if (bef.size() == 0) return;
    ans[(*bef.begin()).second] = i;
    bef.erase(bef.begin());
  }
  add(ans);
}
void gridi2() {
  for (int i = 1; i <= n; i++)
    lt[i] = n + 1 - lt[i], rt[i] = n + 1 - rt[i], swap(lt[i], rt[i]);
  vector<int> ans(n);
  set<pair<pair<int, int>, int> > st;
  set<pair<int, int> > bef;
  for (int i = 1; i <= n; i++) st.insert({{lt[i], rt[i]}, i - 1});
  for (int i = 1; i <= n; i++) {
    while ((st.size() > 0) and ((*st.begin()).first.first <= i)) {
      bef.insert({(*st.begin()).first.second, (*st.begin()).second});
      st.erase(st.begin());
    }
    if (bef.size() == 0) {
      for (int i = 1; i <= n; i++)
        lt[i] = n + 1 - lt[i], rt[i] = n + 1 - rt[i], swap(lt[i], rt[i]);
      return;
    }
    ans[(*bef.begin()).second] = i;
    bef.erase(bef.begin());
  }
  for (int i = 1; i <= n; i++)
    lt[i] = n + 1 - lt[i], rt[i] = n + 1 - rt[i], swap(lt[i], rt[i]);
  for (int& i : ans) i = n + 1 - i;
  add(ans);
}
void gridi3() {
  vector<int> ans(n);
  set<pair<pair<int, int>, int> > st;
  set<pair<int, int> > bef;
  for (int i = 1; i <= n; i++) st.insert({{lt[i], rt[i]}, -i + 1});
  for (int i = 1; i <= n; i++) {
    while ((st.size() > 0) and ((*st.begin()).first.first <= i)) {
      bef.insert({(*st.begin()).first.second, (*st.begin()).second});
      st.erase(st.begin());
    }
    if (bef.size() == 0) return;
    ans[-(*bef.begin()).second] = i;
    bef.erase(bef.begin());
  }
  add(ans);
}
void gridi4() {
  for (int i = 1; i <= n; i++)
    lt[i] = n + 1 - lt[i], rt[i] = n + 1 - rt[i], swap(lt[i], rt[i]);
  vector<int> ans(n);
  set<pair<pair<int, int>, int> > st;
  set<pair<int, int> > bef;
  for (int i = 1; i <= n; i++) st.insert({{lt[i], rt[i]}, -i + 1});
  for (int i = 1; i <= n; i++) {
    while ((st.size() > 0) and ((*st.begin()).first.first <= i)) {
      bef.insert({(*st.begin()).first.second, (*st.begin()).second});
      st.erase(st.begin());
    }
    if (bef.size() == 0) {
      for (int i = 1; i <= n; i++)
        lt[i] = n + 1 - lt[i], rt[i] = n + 1 - rt[i], swap(lt[i], rt[i]);
      return;
    }
    ans[-(*bef.begin()).second] = i;
    bef.erase(bef.begin());
  }
  for (int i = 1; i <= n; i++)
    lt[i] = n + 1 - lt[i], rt[i] = n + 1 - rt[i], swap(lt[i], rt[i]);
  for (int& i : ans) i = n + 1 - i;
  add(ans);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> lt[i] >> rt[i];
  n = m = N;
  gridi();
  gridi2();
  gridi3();
  gridi4();
  while (clock() * 1.0 / CLOCKS_PER_SEC < 0.5) try_();
  cout << "YES\n";
  if (answers.empty()) cout << 1 / 0;
  for (int i : (*answers.begin())) cout << i << " ";
  cout << "\n";
  return 0;
}
