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 N = 2e3 + 10;
inline void read(int &a) {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  a = x * f;
}
int n, root, a[N], ff, fa[N], to[N * 2], head[N], nxt[N * 2], ct, num = 0,
                                                                  ans[N];
void addedge(int x, int y) {
  nxt[++ct] = head[x];
  head[x] = ct;
  to[ct] = y;
}
vector<pair<int, int> > dfs(int u) {
  vector<pair<int, int> > tmp;
  if (ff) return tmp;
  bool f = 1;
  for (int i = head[u]; i; i = nxt[i]) {
    int y = to[i];
    if (y == fa[u]) continue;
    f = 0;
    vector<pair<int, int> > cur = dfs(y);
    for (auto x : cur) tmp.push_back(x);
  }
  if (f) {
    if (a[u]) {
      ff = 1;
      return tmp;
    }
    tmp.push_back(make_pair(++num, u));
  } else {
    if (a[u] == tmp.size()) {
      tmp.push_back(make_pair(++num, u));
      return tmp;
    }
    if (a[u] > tmp.size()) {
      ff = 1;
      return tmp;
    }
    sort(tmp.begin(), tmp.end());
    for (int i = a[u]; i <= (int)(int)tmp.size() - 1; ++i) {
      tmp[i].first += 2;
    }
    num += 2;
    int nw = tmp[a[u]].first - 1;
    tmp.push_back(make_pair(nw, u));
  }
  return tmp;
}
int main() {
  cin >> n;
  for (int i = 1; i <= (int)n; ++i) {
    cin >> fa[i] >> a[i];
    if (!fa[i]) {
      root = i;
      continue;
    }
    addedge(fa[i], i);
    addedge(i, fa[i]);
  }
  vector<pair<int, int> > an = dfs(root);
  for (auto x : an) {
    ans[x.second] = x.first;
  }
  if (ff)
    puts("NO");
  else {
    puts("YES");
    for (int i = 1; i <= (int)n; ++i) cout << ans[i] << " ";
  }
  return 0;
}
