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>
using namespace std;
const int MAXN = 200010;
int _w;
int n, m, k, r[MAXN];
int minr[MAXN], maxr[MAXN];
namespace G {
int head[MAXN], nxt[MAXN], to[MAXN], eid;
void init() {
  eid = 0;
  memset(head, -1, sizeof head);
}
void adde(int u, int v) { to[eid] = v, nxt[eid] = head[u], head[u] = eid++; }
}  // namespace G
int topo[MAXN];
bool toposort() {
  using namespace G;
  static int in[MAXN];
  static queue<int> q;
  for (int u = 1; u <= n; ++u)
    for (int i = head[u]; ~i; i = nxt[i]) ++in[to[i]];
  for (int u = 1; u <= n; ++u)
    if (!in[u]) q.push(u);
  for (int i = 1; i <= n; ++i) {
    if (q.empty()) return false;
    int u = q.front();
    q.pop();
    topo[i] = u;
    for (int e = head[u]; ~e; e = nxt[e]) {
      int v = to[e];
      if (!--in[v]) q.push(v);
    }
  }
  return true;
}
struct Node {
  int u;
  Node(int u) : u(u) {}
  bool operator<(const Node &rhs) const { return maxr[u] > maxr[rhs.u]; }
  operator int() const { return u; }
};
bool solve() {
  using namespace G;
  static priority_queue<Node> pq;
  static vector<int> vec[MAXN];
  for (int i = 1; i <= n; ++i)
    if (r[i])
      minr[i] = maxr[i] = r[i];
    else
      minr[i] = 1, maxr[i] = k;
  for (int i = 1; i <= n; ++i) {
    int u = topo[i];
    for (int e = head[u]; ~e; e = nxt[e]) {
      int v = to[e];
      maxr[v] = min(maxr[v], maxr[u] - 1);
    }
  }
  for (int i = n; i >= 1; --i) {
    int u = topo[i];
    for (int e = head[u]; ~e; e = nxt[e]) {
      int v = to[e];
      minr[u] = max(minr[u], minr[v] + 1);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (minr[i] > maxr[i]) return false;
    vec[minr[i]].push_back(i);
  }
  for (int rk = 1; rk <= k; ++rk) {
    for (vector<int>::iterator it = vec[rk].begin(); it != vec[rk].end(); ++it)
      pq.push(*it);
    if (pq.empty()) return false;
    int u = pq.top();
    pq.pop();
    r[u] = rk;
    while (!pq.empty() && maxr[pq.top()] == rk) r[pq.top()] = rk, pq.pop();
  }
  return true;
}
int main() {
  _w = scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) _w = scanf("%d", r + i);
  G::init();
  for (int i = 0; i < m; ++i) {
    int u, v;
    _w = scanf("%d%d", &u, &v);
    G::adde(u, v);
  }
  if (!toposort()) return puts("-1"), 0;
  if (!solve()) {
    puts("-1");
  } else {
    for (int i = 1; i <= n; ++i) printf("%d ", r[i]);
    puts("");
  }
  return 0;
}