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>
using namespace std;
const int inf = (int)1.01e9;
const long long infll = (long long)1.01e18;
const long double eps = 1e-9;
const long double pi = acos((long double)-1);
mt19937 mrand(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int x) { return mrand() % x; }
void precalc() {}
const int maxn = 2005;
int n;
int c[maxn];
vector<int> g[maxn];
int root;
bool read() {
  if (scanf("%d", &n) < 1) {
    return false;
  }
  for (int i = 0; i < n; i++) {
    g[i].clear();
  }
  root = -1;
  for (int i = 0; i < n; i++) {
    int p;
    scanf("%d%d", &p, &c[i]);
    p--;
    if (p == -1) {
      root = i;
    } else {
      g[p].push_back(i);
    }
  }
  return true;
}
int ts[maxn];
int a[maxn];
vector<int> vs[maxn];
bool dfs(int v) {
  ts[v] = 1;
  for (int i = 0; i < ((int)(g[v]).size()); i++) {
    int u = g[v][i];
    if (!dfs(u)) {
      return false;
    }
    ts[v] += ts[u];
  }
  if (c[v] >= ts[v]) {
    return false;
  }
  vs[v].clear();
  int k = 0;
  for (int i = 0; i < ((int)(g[v]).size()); i++) {
    int u = g[v][i];
    for (int j = 0; j < ((int)(vs[u]).size()); j++) {
      int w = vs[u][j];
      a[w] += k;
      vs[v].push_back(w);
    }
    vs[u].clear();
    k += ts[u];
  }
  for (int i = 0; i < ((int)(vs[v]).size()); i++) {
    int u = vs[v][i];
    if (a[u] >= c[v]) {
      a[u]++;
    }
  }
  a[v] = c[v];
  vs[v].push_back(v);
  return true;
}
void solve() {
  if (!dfs(root)) {
    printf("NO\n");
    return;
  }
  printf("YES\n");
  for (int i = 0; i < n; i++) {
    printf("%d ", a[i] + 1);
  }
  printf("\n");
}
int main() {
  precalc();
  while (read()) {
    solve();
  }
  return 0;
}
