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>
const int maxn = 200000 + 5;
const int maxm = 200000 + 5;
std::vector<int> nG[maxn];
std::vector<int> rG[maxn];
void add_edge(int u, int v, bool rev = 0) {
  auto &G = (rev ? rG : nG);
  G[u].push_back(v);
}
void init_edge() {
  for (int i = 0; i < maxn; i++) {
    nG[i].clear();
    rG[i].clear();
  }
}
int n, m, k;
int in[maxn];
int min_rank[maxn], max_rank[maxn], r[maxn];
struct Node {
  int min_rank;
  int id;
  const bool operator<(const Node &rhs) const {
    return min_rank < rhs.min_rank;
  }
};
std::vector<Node> s[maxn];
std::vector<int> topo;
void toposort() {
  std::queue<int> Q;
  for (int i = 1; i <= n; i++) {
    if (!in[i]) {
      Q.push(i);
      topo.push_back(i);
    }
  }
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int i = 0; i < nG[u].size(); i++) {
      int v = nG[u][i];
      in[v]--;
      if (!in[v]) {
        Q.push(v);
        topo.push_back(v);
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", r + i);
    max_rank[i] = r[i] ? r[i] : k;
    min_rank[i] = r[i] ? r[i] : 1;
  }
  init_edge();
  int u, v;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    in[v]++;
    add_edge(v, u, 1);
  }
  toposort();
  if (topo.size() != n) {
    puts("-1");
    return 0;
  }
  for (int i = 0; i < topo.size(); i++) {
    u = topo[i];
    for (int j = 0; j < nG[u].size(); j++) {
      v = nG[u][j];
      max_rank[v] = std::min(max_rank[v], max_rank[u] - 1);
    }
  }
  for (int i = topo.size() - 1; i >= 0; i--) {
    u = topo[i];
    for (int j = 0; j < rG[u].size(); j++) {
      v = rG[u][j];
      min_rank[v] = std::max(min_rank[v], min_rank[u] + 1);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (max_rank[i] < min_rank[i]) {
      puts("-1");
      return 0;
    } else {
      s[max_rank[i]].push_back((Node){min_rank[i], i});
    }
  }
  std::priority_queue<Node> Q;
  for (int i = k; i > 0; i--) {
    for (int j = 0; j < s[i].size(); j++) {
      Q.push(s[i][j]);
    }
    if (Q.empty()) {
      puts("-1");
      return 0;
    }
    int x = Q.top().id;
    Q.pop();
    r[x] = i;
    while (!Q.empty() && Q.top().min_rank == i) {
      r[Q.top().id] = i;
      Q.pop();
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", r[i]);
  }
  printf("\n");
}