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 N = 200001;
const int oo = 1000000000;
int n, m, k, indeg[N], rdeg[N], fr[N], rankk[N], ans[N], is[N], al[N];
vector<int> g[N], p[N];
void no() {
  cout << -1 << endl;
  exit(0);
}
int main() {
  cin >> n >> m >> k;
  if (k > n) no();
  for (int i = 0; i < n; ++i) {
    scanf("%d", &rankk[i]);
    if (rankk[i]) al[i] = 1;
  }
  for (int a, b, i = 0; i < m; ++i) {
    scanf("%d%d", &a, &b);
    --a;
    --b;
    ++indeg[b];
    ++rdeg[a];
    g[a].push_back(b);
    p[b].push_back(a);
  }
  queue<int> q;
  for (int i = 0; i < n; ++i)
    if (!indeg[i]) q.push(i);
  int ls = k, cttt = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    ++cttt;
    int rk = rankk[u];
    if (!rk) {
      rk = k;
      for (int i = 0; i < p[u].size(); ++i) rk = min(rk, rankk[p[u][i]] - 1);
    }
    rankk[u] = rk;
    is[rk] = 1;
    ls = min(ls, rk);
    for (int i = 0; i < g[u].size(); ++i) {
      if (rankk[g[u][i]] >= rk) no();
      if (!--indeg[g[u][i]]) q.push(g[u][i]);
    }
  }
  if (cttt < n || ls < 1) no();
  set<int> st;
  set<pair<int, int> > oth;
  for (int i = 1; i <= k; ++i)
    if (!is[i]) st.insert(i);
  priority_queue<pair<int, int> > qq;
  for (int i = 0; i < n; ++i) {
    ++fr[rankk[i]];
    if (!rdeg[i]) qq.push(pair<int, int>(-rankk[i], i));
  }
  while (!qq.empty() && !st.empty()) {
    int u = qq.top().second;
    qq.pop();
    bool fff = 1;
    int d = *st.begin();
    if (!al[u] && rankk[u] > d) {
      st.erase(st.begin());
      if (!--fr[rankk[u]]) st.insert(rankk[u]);
      ++fr[d];
      rankk[u] = d;
    } else if (al[u] && rankk[u] > d)
      oth.insert(pair<int, int>(rankk[u], u)), fff = 0;
    if (fff)
      for (int i = 0; i < p[u].size(); ++i)
        if (!--rdeg[p[u][i]]) qq.push(pair<int, int>(-rankk[p[u][i]], p[u][i]));
    while (!oth.empty() && !st.empty() && oth.begin()->first < *st.begin()) {
      qq.push(pair<int, int>(-oth.begin()->first, oth.begin()->second));
      oth.erase(oth.begin());
    }
  }
  if (st.size()) no();
  for (int i = 0; i < n; ++i) printf("%d ", rankk[i]);
  puts("");
}