Problem: There are n military men in the Berland army. Some of them have given orders to other military men by now. Given m pairs (xi, yi), meaning that the military man xi gave the i-th order to another military man yi.

It is time for reform! The Berland Ministry of Defence plans to introduce ranks in the Berland army. Each military man should be assigned a rank — integer number between 1 and k, inclusive. Some of them have been already assigned a rank, but the rest of them should get a rank soon.

Help the ministry to assign ranks to the rest of the army so that:

  * for each of m orders it is true that the rank of a person giving the order (military man xi) is strictly greater than the rank of a person receiving the order (military man yi); 
  * for each rank from 1 to k there is at least one military man with this rank. 

Input

The first line contains three integers n, m and k (1 ≤ n ≤ 2·105, 0 ≤ m ≤ 2·105, 1 ≤ k ≤ 2·105) — number of military men in the Berland army, number of orders and number of ranks.

The second line contains n integers r1, r2, ..., rn, where ri > 0 (in this case 1 ≤ ri ≤ k) means that the i-th military man has been already assigned the rank ri; ri = 0 means the i-th military man doesn't have a rank yet.

The following m lines contain orders one per line. Each order is described with a line containing two integers xi, yi (1 ≤ xi, yi ≤ n, xi ≠ yi). This line means that the i-th order was given by the military man xi to the military man yi. For each pair (x, y) of military men there could be several orders from x to y.

Output

Print n integers, where the i-th number is the rank of the i-th military man. If there are many solutions, print any of them.

If there is no solution, print the only number -1.

Examples

Input

5 3 3
0 3 0 0 2
2 4
3 4
3 5


Output

1 3 3 2 2 


Input

7 6 5
0 4 5 4 1 0 0
6 1
3 6
3 1
7 5
7 1
7 4


Output

2 4 5 4 1 3 5 


Input

2 2 2
2 1
1 2
2 1


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;
int IT_MAX = 1 << 18;
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x1f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-12;
vector<int> conn[200050];
vector<int> rconn[200050];
int inc[200050];
int mx[200050];
int ans[200050];
int mn[200050];
set<pair<int, int> > Sx1;
set<pair<int, int> > Sx2;
vector<int> Vl;
void propagate() {
  while (!Vl.empty()) {
    int t = Vl.back();
    Vl.pop_back();
    for (auto it : rconn[t]) {
      mn[it] = max(mn[it], ans[t] + 1);
      if (--inc[it] == 0) Sx2.emplace(mn[it], it);
    }
  }
}
int main() {
  int N, M, K, i, j;
  scanf("%d %d %d", &N, &M, &K);
  for (i = 1; i <= N; i++) {
    int t;
    scanf("%d", &t);
    ans[i] = t;
    if (t != 0)
      mx[i] = mn[i] = t;
    else {
      mn[i] = 1;
      mx[i] = K;
    }
  }
  for (i = 1; i <= M; i++) {
    int t1, t2;
    scanf("%d %d", &t1, &t2);
    conn[t1].push_back(t2);
    rconn[t2].push_back(t1);
  }
  for (i = 1; i <= N; i++) inc[i] = rconn[i].size();
  vector<int> Vu;
  for (i = 1; i <= N; i++)
    if (!inc[i]) Vu.push_back(i);
  for (i = 0; i < Vu.size(); i++) {
    for (auto it : conn[Vu[i]]) {
      mx[it] = min(mx[it], mx[Vu[i]] - 1);
      if (--inc[it] == 0) Vu.push_back(it);
    }
  }
  if (Vu.size() != N) return !printf("-1\n");
  Vu.clear();
  for (i = 1; i <= N; i++) {
    if (mx[i] <= 0) return !printf("-1\n");
    if (ans[i] && ans[i] > mx[i]) return !printf("-1\n");
  }
  for (i = 1; i <= N; i++) inc[i] = conn[i].size();
  for (i = 1; i <= N; i++)
    if (inc[i] == 0) Sx2.emplace(mn[i], i);
  for (i = 1; i <= K; i++) {
    while (!Sx2.empty()) {
      if (Sx2.begin()->first != i) break;
      int t = Sx2.begin()->second;
      Sx2.erase(Sx2.begin());
      Sx1.emplace(mx[t], t);
    }
    if (Sx1.empty()) return !printf("-1\n");
    auto it = Sx1.begin();
    Vl.push_back(it->second);
    ans[it->second] = i;
    Sx1.erase(it);
    while (!Sx1.empty()) {
      if (Sx1.begin()->first != i) break;
      int t = Sx1.begin()->second;
      Sx1.erase(Sx1.begin());
      ans[t] = i;
      Vl.push_back(t);
    }
    propagate();
  }
  for (i = 1; i <= N; i++) printf("%d ", ans[i]);
  return !printf("\n");
}