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;
using i64 = long long;
const i64 MOD = 1e9 + 7;
const i64 INF = 1e18 + 7;
template <typename T>
bool chmin(T& x, T y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
template <typename T>
bool chmax(T& x, T y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
template <typename T>
struct Compression {
  vector<T> compvec;
  Compression(vector<T>& inp) {
    compvec = inp;
    sort(compvec.begin(), compvec.end());
    compvec.erase(unique(compvec.begin(), compvec.end()), compvec.end());
  }
  i64 Index(T val) {
    auto it = lower_bound(compvec.begin(), compvec.end(), val);
    return distance(compvec.begin(), it);
  }
  vector<T>& operator*() { return compvec; }
};
vector<i64> par;
vector<i64> c, a;
vector<vector<i64>> childs;
int n, cnt;
vector<i64> h(i64 x) {
  vector<i64> ret;
  for (auto& y : childs[x]) {
    ret.emplace_back(a[y]);
    auto r = h(y);
    if (ret.size() < r.size()) swap(ret, r);
    for (auto& elm : r) ret.emplace_back(elm);
  }
  return ret;
}
void p(i64 x, i64 val) {
  if (a[x] >= val) ++a[x];
  for (auto& y : childs[x]) p(y, val);
}
void g(i64 x) {
  i64 k = c[x];
  auto elm = h(x);
  if (elm.size() < k) {
    cout << "NO" << endl;
    exit(0);
  }
  nth_element(elm.begin(), next(elm.begin(), k), elm.end());
  elm.emplace_back(elm.empty() ? cnt + 1
                               : *max_element(elm.begin(), elm.end()) + 1);
  i64 val = elm[k];
  for (auto& y : childs[x]) p(y, val);
  a[x] = val;
}
void f(i64 x) {
  for (auto& y : childs[x]) f(y);
  ++cnt;
  g(x);
}
signed main() {
  cin >> n;
  par.resize(n);
  c.resize(n);
  childs.resize(n);
  cnt = 0;
  for (i64 i = 0; i < n; ++i) {
    cin >> par[i] >> c[i];
    --par[i];
    if (par[i] != -1) childs[par[i]].emplace_back(i);
  }
  a.assign(n, 0);
  for (i64 i = 0; i < n; ++i)
    if (par[i] == -1) f(i);
  cout << "YES" << endl;
  Compression<i64> comp(a);
  for (i64 i = 0; i < n; ++i) cout << comp.Index(a[i]) + 1 << " \n"[i == n - 1];
}
