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 = 3e5 + 5;
int n, m, k;
int rnk[N];
int arr[N];
int low[N];
vector<int> v[N];
vector<int> rv[N];
vector<int> topo;
int indeg[N];
void toposort() {
  memset(indeg, 0, sizeof(indeg));
  for (int i = 1; i <= n; ++i) {
    for (int x : v[i]) {
      ++indeg[x];
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (!indeg[i]) {
      v[0].emplace_back(i);
      rv[i].emplace_back(0);
      ++indeg[i];
    }
  }
  topo.clear();
  queue<int> que;
  while (!que.empty()) {
    que.pop();
  }
  que.push(0);
  while (!que.empty()) {
    int node = que.front();
    topo.emplace_back(node);
    que.pop();
    for (int next : v[node]) {
      --indeg[next];
      if (!indeg[next]) {
        que.push(next);
      }
    }
  }
}
vector<int> there[N];
set<int> s;
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", rnk + i);
  }
  for (int i = 1; i <= m; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    v[a].emplace_back(b);
    rv[b].emplace_back(a);
  }
  ++k;
  rnk[0] = k;
  toposort();
  if (topo.size() != n + 1) {
    printf("-1\n");
    return 0;
  }
  for (int i = n; i >= 0; --i) {
    int node = topo[i];
    low[node] = 1;
    for (int next : v[node]) {
      low[node] = max(low[node], low[next] + 1);
    }
    if (rnk[node] != 0) {
      if (rnk[node] < low[node]) {
        printf("-1\n");
        return 0;
      }
      low[node] = rnk[node];
    }
    arr[node] = low[node];
    there[arr[node]].emplace_back(node);
    if (arr[node] > k) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = k; i >= 1; --i) {
    s.insert(i);
  }
  for (int i = 0; i <= n; ++i) {
    s.erase(arr[i]);
  }
  for (int i = k; i >= 1; --i) {
    vector<pair<int, int> > g;
    g.clear();
    for (int x : there[i]) {
      int val = k;
      for (int y : rv[x]) {
        val = min(val, arr[y] - 1);
      }
      if (rnk[x] != 0) {
        val = rnk[x];
      }
      g.emplace_back(make_pair(val, x));
    }
    s.insert(i);
    sort(g.begin(), g.end());
    reverse(g.begin(), g.end());
    for (auto itt : g) {
      int node = itt.second;
      int high = itt.first;
      if (low[node] > high) {
        printf("-1\n");
      }
      auto it = s.lower_bound(high + 1);
      int val = -1;
      if (it != s.begin()) {
        --it;
        val = *it;
      }
      if (val >= low[node]) {
        arr[node] = val;
      } else {
        arr[node] = high;
      }
      s.erase(arr[node]);
    }
  }
  if (!s.empty()) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d%c", arr[i], " \n"[i == n]);
  }
}