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;
long long mod = 1000000007LL;
long long large = 1000000000000000000LL;
vector<int> dfs_num, dfs_low, S, visited;
vector<vector<int> > adj;
vector<vector<int> > Scc;
int dfsNumberCounter = 0, numSCC = 0;
void tarjanSCC(int u) {
  dfs_low[u] = dfs_num[u] = dfsNumberCounter++;
  S.push_back(u);
  visited[u] = 1;
  for (int j = 0; j < (int)adj[u].size(); j++) {
    int v = adj[u][j];
    if (dfs_num[v] == -1) {
      tarjanSCC(v);
    }
    if (visited[v] == 1) {
      dfs_low[u] = min(dfs_low[u], dfs_low[v]);
    }
  }
  if (dfs_low[u] == dfs_num[u]) {
    Scc.push_back(vector<int>(0, 0));
    while (true) {
      int v = S.back();
      S.pop_back();
      visited[v] = 0;
      Scc[numSCC].push_back(v);
      if (u == v) break;
    }
    numSCC++;
  }
}
int uv, up;
vector<int> mini;
void update(int o, int l, int r) {
  if (l == r) {
    mini[o] = uv;
  } else {
    int m = (l + r) / 2;
    if (up <= m) update(o * 2, l, m);
    if (up > m) update(o * 2 + 1, m + 1, r);
    mini[o] = min(mini[o * 2], mini[o * 2 + 1]);
  }
}
int ql, qr, re;
void query(int o, int l, int r) {
  if (ql <= l && qr >= r) {
    re = min(re, mini[o]);
  } else {
    int m = (l + r) / 2;
    if (ql <= m) query(o * 2, l, m);
    if (qr > m) query(o * 2 + 1, m + 1, r);
  }
}
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  k--;
  dfs_num.assign(n, -1);
  dfs_low.assign(n, 0);
  visited.assign(n, 0);
  adj.assign(n, vector<int>(0, 0));
  vector<int> a(n, 0);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  vector<int> b = a;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
  }
  for (int i = 0; i < n; i++) {
    if (dfs_num[i] == -1) {
      tarjanSCC(i);
    }
  }
  bool flag = true;
  for (int i = 0; i < numSCC; i++) {
    if (Scc[i].size() > 1) flag = false;
  }
  if (!flag) {
    cout << -1 << endl;
    return 0;
  }
  vector<vector<int> > in(n, vector<int>(0, 0));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < (int)adj[i].size(); j++) {
      int v = adj[i][j];
      in[v].push_back(i);
    }
  }
  vector<int> deg(n, 0);
  for (int i = 0; i < n; i++) deg[i] = (int)in[i].size();
  set<int> s;
  for (int i = 0; i < n; i++)
    if (deg[i] == 0) {
      s.insert(i);
    }
  while (!s.empty()) {
    int u = *s.begin();
    s.erase(u);
    if (a[u] != -1) {
      for (int j = 0; j < (int)in[u].size(); j++) {
        int from = in[u][j];
        if (a[u] >= a[from]) flag = false;
      }
      if (!flag) break;
    } else {
      a[u] = k;
      for (int j = 0; j < (int)in[u].size(); j++) {
        int from = in[u][j];
        a[u] = min(a[u], a[from] - 1);
      }
    }
    for (int j = 0; j < (int)adj[u].size(); j++) {
      int v = adj[u][j];
      deg[v]--;
      if (deg[v] == 0) s.insert(v);
    }
  }
  for (int i = 0; i < n; i++)
    if (a[i] < 0) flag = false;
  if (!flag) {
    cout << "-1" << endl;
    return 0;
  }
  mini.assign(4 * k + 10, 0);
  vector<int> cnt(k + 1, 0);
  for (int i = 0; i < n; i++) cnt[a[i]]++;
  for (int i = 0; i <= k; i++) {
    up = i;
    uv = i;
    if (cnt[i] != 0) uv = 1000000;
    update(1, 0, k);
  }
  vector<pair<int, int> > e;
  for (int i = 0; i < n; i++) e.push_back(pair<int, int>(a[i], i));
  sort(e.begin(), e.end());
  for (int i = 0; i < n; i++) {
    int t = e[i].second;
    if (b[t] != -1) continue;
    ql = 0;
    for (int j = 0; j < (int)adj[t].size(); j++) {
      int v = adj[t][j];
      ql = max(ql, a[v] + 1);
    }
    qr = a[t] - 1;
    if (qr < ql) continue;
    re = 1000000;
    query(1, 0, k);
    if (re == 1000000) continue;
    cnt[a[t]]--;
    if (cnt[a[t]] == 0) {
      up = a[t];
      uv = a[t];
      update(1, 0, k);
    }
    a[t] = re;
    cnt[re]++;
    up = re;
    uv = 1000000;
    update(1, 0, k);
  }
  if (mini[1] != 1000000) {
    cout << "-1" << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) cout << a[i] + 1 << " ";
  cout << endl;
  return 0;
}