#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long qpow(long long a, long long b, long long m) {
  long long r = 1;
  a %= m;
  for (; b; b >>= 1) {
    if (b & 1) r = r * a % m;
    a = a * a % m;
  }
  return r;
}
const int inf = 0x7fffffff;
const int maxn = 2e3 + 10;
vector<int> g[maxn];
int cnt[maxn];
int c[maxn];
int n, rt;
void dfs0(int u) {
  cnt[u] = 1;
  for (int v : g[u]) {
    dfs0(v);
    cnt[u] += cnt[v];
  }
}
vector<int> ve[maxn];
int xx = 3000, tt = 0;
int val[maxn];
map<int, int> mp;
void dfs(int u) {
  for (int v : g[u]) {
    dfs(v);
    for (int x : ve[v]) ve[u].push_back(x);
  }
  if ((int)(g[u]).size()) {
    vector<int>::iterator it = ve[u].begin();
    for (int i = 0; i < c[u]; i++) it++;
    ve[u].insert(it, 1, u);
  } else
    ve[u].push_back(u);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p >> c[i];
    if (!p)
      rt = i;
    else
      g[p].push_back(i);
  }
  dfs0(rt);
  for (int i = 1; i <= n; i++)
    if (c[i] >= cnt[i]) {
      cout << "NO" << endl;
      return 0;
    }
  dfs(rt);
  cout << "YES" << endl;
  int ttt = 0;
  for (int i = 0; i < n; i++) {
    val[ve[rt][i]] = ++ttt;
  }
  for (int i = 1; i <= n; i++) {
    if (i - 1) cout << " ";
    cout << val[i];
  }
  cout << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long qpow(long long a, long long b, long long m) {
  long long r = 1;
  a %= m;
  for (; b; b >>= 1) {
    if (b & 1) r = r * a % m;
    a = a * a % m;
  }
  return r;
}
const int inf = 0x7fffffff;
const int maxn = 2e3 + 10;
vector<int> g[maxn];
int cnt[maxn];
int c[maxn];
int n, rt;
void dfs0(int u) {
  cnt[u] = 1;
  for (int v : g[u]) {
    dfs0(v);
    cnt[u] += cnt[v];
  }
}
vector<int> ve[maxn];
int xx = 3000, tt = 0;
int val[maxn];
map<int, int> mp;
void dfs(int u) {
  for (int v : g[u]) {
    dfs(v);
    for (int x : ve[v]) ve[u].push_back(x);
  }
  if ((int)(g[u]).size()) {
    vector<int>::iterator it = ve[u].begin();
    for (int i = 0; i < c[u]; i++) it++;
    ve[u].insert(it, 1, u);
  } else
    ve[u].push_back(u);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p >> c[i];
    if (!p)
      rt = i;
    else
      g[p].push_back(i);
  }
  dfs0(rt);
  for (int i = 1; i <= n; i++)
    if (c[i] >= cnt[i]) {
      cout << "NO" << endl;
      return 0;
    }
  dfs(rt);
  cout << "YES" << endl;
  int ttt = 0;
  for (int i = 0; i < n; i++) {
    val[ve[rt][i]] = ++ttt;
  }
  for (int i = 1; i <= n; i++) {
    if (i - 1) cout << " ";
    cout << val[i];
  }
  cout << endl;
  return 0;
}

```