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 = 200005;
int r[MAXN];
vector<int> hui[MAXN], backward[MAXN];
void Fail() {
  cout << -1 << endl;
  exit(0);
}
int state[MAXN];
int topo[MAXN];
void Toposort(int v) {
  static int pntTopo = 1;
  state[v] = 1;
  for (int u : hui[v]) {
    if (state[u] == 1) {
      Fail();
    }
    if (state[u] == 0) {
      Toposort(u);
    }
  }
  topo[pntTopo++] = v;
  state[v] = 2;
}
struct Node {
  int l, r, idx;
};
bool compare(Node A, Node B) { return A.l < B.l; }
Node nodes[MAXN];
int ans[MAXN];
priority_queue<pair<int, int>> Q;
int main() {
  int N, M, K;
  cin >> N >> M >> K;
  for (int i = 1; i <= N; i++) {
    cin >> r[i];
  }
  for (int i = 1; i <= M; i++) {
    int x, y;
    cin >> x >> y;
    hui[x].push_back(y);
    backward[y].push_back(x);
  }
  for (int v = 1; v <= N; v++) {
    if (state[v] == 0) {
      Toposort(v);
    }
    nodes[v].idx = v;
  }
  for (int i = 1; i <= N; i++) {
    int v = topo[i];
    int Max = 0;
    for (int u : hui[v]) {
      Max = max(Max, nodes[u].l);
    }
    if (r[v] != 0) {
      if (Max >= r[v]) {
        Fail();
      }
      nodes[v].l = r[v];
    } else {
      nodes[v].l = Max + 1;
    }
  }
  for (int i = N; i >= 1; i--) {
    int v = topo[i];
    int Min = K + 1;
    for (int u : backward[v]) {
      Min = min(Min, nodes[u].r);
    }
    if (r[v] != 0) {
      if (Min <= r[v]) {
        Fail();
      }
      nodes[v].r = r[v];
    } else {
      nodes[v].r = Min - 1;
    }
  }
  for (int v = 1; v <= N; v++) {
    if (nodes[v].r < nodes[v].l) {
      Fail();
    }
  }
  sort(nodes + 1, nodes + N + 1, compare);
  int curr = 1;
  int pntNodes = 1;
  for (int i = 1; i <= N; i++) {
    while (pntNodes <= N && nodes[pntNodes].l == curr) {
      Q.push({-nodes[pntNodes].r, nodes[pntNodes].idx});
      pntNodes++;
    }
    if (Q.empty()) {
      Fail();
    }
    pair<int, int> x = Q.top();
    Q.pop();
    if (-x.first >= curr) {
      ans[x.second] = curr++;
    } else {
      ans[x.second] = -x.first;
    }
  }
  if (curr != K + 1) {
    Fail();
  }
  for (int i = 1; i <= N; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}