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 = 2e5 + 100;
long long n, m, k;
long long pre[maxn];
vector<long long> g[maxn], rg[maxn];
long long r[maxn];
long long dat[maxn][2];
bool prepro() {
  for (int i = 0; i < n; ++i) {
    if (r[i] == 0)
      dat[i][0] = dat[i][1] = -1;
    else
      dat[i][0] = dat[i][1] = r[i];
  }
  queue<int> q;
  for (int i = 0; i < n; ++i) {
    pre[i] = rg[i].size();
    if (pre[i] == 0) q.push(i);
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (dat[u][0] == -1) {
      dat[u][0] = 1;
      for (int j = 0; j < rg[u].size(); ++j)
        dat[u][0] = max(dat[u][0], dat[rg[u][j]][0] + 1);
    }
    for (int i = 0; i < g[u].size(); ++i) {
      int v = g[u][i];
      --pre[v];
      if (pre[v] == 0) q.push(v);
    }
  }
  for (int i = 0; i < n; ++i) {
    pre[i] = g[i].size();
    if (pre[i] == 0) q.push(i);
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (dat[u][1] == -1) {
      dat[u][1] = k;
      for (int j = 0; j < g[u].size(); ++j)
        dat[u][1] = min(dat[u][1], dat[g[u][j]][1] - 1);
    }
    for (int i = 0; i < rg[u].size(); ++i) {
      int v = rg[u][i];
      --pre[v];
      if (pre[v] == 0) q.push(v);
    }
  }
  for (int i = 0; i < n; ++i) {
    if (dat[i][0] == -1 or dat[i][1] == -1) return false;
    for (int j = 0; j < g[i].size(); ++j) {
      int v = g[i][j];
    }
  }
  return true;
}
bool solve() {
  priority_queue<pair<long long, long long> > q1, q2;
  for (int i = 0; i < n; ++i) {
    q1.push(pair<long long, long long>(-dat[i][0], i));
  }
  for (int i = 1; i <= k; ++i) {
    while (!q1.empty() and q1.top().first == -i) {
      pair<long long, long long> aux = q1.top();
      q1.pop();
      int j = aux.second;
      q2.push(pair<long long, long long>(-dat[j][1], j));
    }
    if (q2.empty()) return false;
    pair<long long, long long> aux = q2.top();
    q2.pop();
    r[aux.second] = i;
    while (!q2.empty() and q2.top().first == -i) {
      r[q2.top().second] = i;
      q2.pop();
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < g[i].size(); ++j) {
      int v = g[i][j];
      if (r[i] >= r[v]) return false;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < n; ++i) cin >> r[i];
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    rg[x].push_back(y);
    g[y].push_back(x);
  }
  if (!prepro()) {
    cout << "-1\n";
    return 0;
  }
  if (!solve()) {
    cout << "-1\n";
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    if (i) cout << ' ';
    cout << r[i];
  }
  cout << '\n';
}