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>
long long dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
long long dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
using namespace std;
class pa3 {
 public:
  long long x;
  long long y, z;
  pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3 &p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3 &p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  long long x;
  long long y, z, w;
  pa4(long long x = 0, long long y = 0, long long z = 0, long long w = 0)
      : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4 &p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  long long x, y;
  pa2(long long x = 0, long long y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2 &p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2 &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2 &p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2 &p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
string itos(long long i) {
  ostringstream s;
  s << i;
  return s.str();
}
long long gcd(long long v, long long b) {
  if (v == 0) return b;
  if (b == 0) return v;
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
long long mod;
long long extgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
pair<long long, long long> operator+(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first + r.first, l.second + r.second};
}
pair<long long, long long> operator-(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first - r.first, l.second - r.second};
}
long long pr[10000100];
long long inv[10000010];
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  wa %= warukazu;
  if (rr % 2 == 1)
    return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
long long comb(long long nn, long long rr) {
  if (rr < 0 || rr > nn || nn < 0) return 0;
  long long r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
void gya(long long ert) {
  pr[0] = 1;
  for (long long i = 1; i <= ert; i++) {
    pr[i] = ((long long)pr[i - 1] * i) % mod;
  }
  inv[ert] = beki(pr[ert], mod - 2, mod);
  for (long long i = ert - 1; i >= 0; i--) {
    inv[i] = (long long)inv[i + 1] * (i + 1) % mod;
  }
}
long long c[2020];
vector<long long> G[2002];
vector<long long> ve[2020];
long long ans[2002];
long long dfs(long long r) {
  long long wa = 1;
  for (auto v : G[r]) wa += dfs(v);
  if (c[r] >= wa) {
    cout << "NO" << endl;
    exit(0);
  }
  long long cnt = 0;
  for (auto v : G[r])
    for (auto w : ve[v]) {
      ve[r].push_back(w);
    }
  ve[r].insert(ve[r].begin() + c[r], r);
  return wa;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  long long ro = -1;
  for (long long i = 1; i <= n; i++) {
    long long y, yy;
    cin >> y >> yy;
    if (y)
      G[y].push_back(i);
    else
      ro = i;
    c[i] = yy;
  }
  dfs(ro);
  long long cnt = 1;
  for (auto v : ve[ro]) {
    ans[v] = cnt;
    cnt++;
  }
  cout << "YES" << endl;
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
