Evlampiy was gifted a rooted tree. The vertices of the tree are numbered from 1 to n. Each of its vertices also has an integer a_i written on it. For each vertex i, Evlampiy calculated c_i — the number of vertices j in the subtree of vertex i, such that a_j < a_i. 

<image>Illustration for the second example, the first integer is a_i and the integer in parentheses is c_i

After the new year, Evlampiy could not remember what his gift was! He remembers the tree and the values of c_i, but he completely forgot which integers a_i were written on the vertices.

Help him to restore initial integers!

Input

The first line contains an integer n (1 ≤ n ≤ 2000) — the number of vertices in the tree.

The next n lines contain descriptions of vertices: the i-th line contains two integers p_i and c_i (0 ≤ p_i ≤ n; 0 ≤ c_i ≤ n-1), where p_i is the parent of vertex i or 0 if vertex i is root, and c_i is the number of vertices j in the subtree of vertex i, such that a_j < a_i.

It is guaranteed that the values of p_i describe a rooted tree with n vertices.

Output

If a solution exists, in the first line print "YES", and in the second line output n integers a_i (1 ≤ a_i ≤ {10}^{9}). If there are several solutions, output any of them. One can prove that if there is a solution, then there is also a solution in which all a_i are between 1 and 10^9.

If there are no solutions, print "NO".

Examples

Input


3
2 0
0 2
2 0


Output


YES
1 2 1 

Input


5
0 1
1 3
2 1
3 0
2 0


Output


YES
2 3 2 1 2
#include <bits/stdc++.h>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
int n, m;
vector<int> sons[2222];
int ASS[2222];
int P[2222];
int C[2222];
int known[2222];
void dfs(int curr, vector<int>& suball, int start = 1) {
  if (!start) suball.push_back(ASS[curr]);
  for (int s : sons[curr]) {
    dfs(s, suball, 0);
  }
}
void dec(int curr, int val, int start = 1) {
  if (!start && ASS[curr] <= val) ASS[curr]--;
  for (int s : sons[curr]) {
    dec(s, val, 0);
  }
}
int cnt(int curr) {
  int ans = 1;
  for (int s : sons[curr]) ans += cnt(s);
  return ans;
}
void inc(int curr, int val = 0, int start = 1) {
  int first = 0;
  if (!start) {
    ASS[curr] += val;
  }
  for (int s : sons[curr]) {
    if (start) {
      if (first) inc(s, first, 0);
      first += cnt(s);
    } else {
      inc(s, val, 0);
    }
  }
}
void fix(int curr, int val) {
  ASS[curr] -= val;
  for (int s : sons[curr]) {
    fix(s, val);
  }
}
int main() {
  scanf("%d", &n);
  int root = 0;
  for (int(i) = 0; (i) < (n); (i)++) {
    scanf("%d%d", P + i, C + i);
    P[i]--;
    if (P[i] == -1)
      root = i;
    else
      sons[P[i]].push_back(i);
  }
  queue<int> Q;
  for (int(i) = 0; (i) < (n); (i)++) {
    if (sons[i].size() == 0) {
      Q.push(i);
    }
  }
  int leafs = 1;
  int mini = 0;
  while (!Q.empty()) {
    int c = Q.front();
    Q.pop();
    if (sons[c].size() == 0) {
      if (C[c] != 0) {
        printf("NO\n");
        return 0;
      }
      ASS[c] = 0;
      known[P[c]]++;
      if (known[P[c]] == sons[P[c]].size()) {
        Q.push(P[c]);
      }
    } else {
      vector<int> suball;
      inc(c);
      dfs(c, suball);
      sort(suball.begin(), suball.end());
      int last = -1000000;
      for (int first : suball) {
        assert(first != last);
        last = first;
      }
      int fixval = suball[0];
      if (C[c] == 0) {
        ASS[c] = suball[0] - 1;
        fixval--;
      } else if (C[c] == suball.size()) {
        ASS[c] = suball[suball.size() - 1] + 1;
      } else if (C[c] < suball.size()) {
        dec(c, suball[C[c] - 1]);
        fixval--;
        ASS[c] = suball[C[c] - 1];
      } else {
        printf("NO\n");
        return 0;
      }
      fix(c, fixval);
      mini = min(mini, ASS[c]);
      if (P[c] != -1) {
        known[P[c]]++;
        if (known[P[c]] == sons[P[c]].size()) {
          Q.push(P[c]);
        }
      }
    }
  }
  printf("YES\n");
  for (int(i) = 0; (i) < (n); (i)++) mini = min(ASS[i], mini);
  for (int(i) = 0; (i) < (n); (i)++) printf("%d ", ASS[i] - mini + 1);
  return 0;
}
