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;
vector<int> e[MAXN], f[MAXN];
int r[MAXN], k;
int mi[MAXN], mx[MAXN], vis[MAXN];
void dfs(int u) {
  if (vis[u] == 1) throw 1;
  if (vis[u] == 2) return;
  vis[u] = 1;
  mi[u] = 1;
  for (auto &v : e[u]) {
    dfs(v);
    mi[u] = max(mi[u], mi[v] + 1);
  }
  if (mi[u] > k) throw 1;
  if (r[u]) {
    if (mi[u] <= r[u])
      mi[u] = r[u];
    else
      throw 1;
  }
  vis[u] = 2;
}
void rdfs(int u) {
  if (vis[u] == 1) throw 1;
  if (vis[u] == 2) return;
  vis[u] = 1;
  mx[u] = k;
  for (auto &v : f[u]) {
    rdfs(v);
    mx[u] = min(mx[u], mx[v] - 1);
  }
  if (mx[u] < 1) throw 1;
  if (r[u]) {
    if (mx[u] >= r[u])
      mx[u] = r[u];
    else
      throw 1;
  }
  vis[u] = 2;
}
void solve(int n) {
  memset(vis, 0, (n + 1) * sizeof(vis[0]));
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  memset(vis, 0, (n + 1) * sizeof(vis[0]));
  for (int i = 1; i <= n; i++)
    if (!vis[i]) rdfs(i);
}
int idx[MAXN];
void remark(int n) {
  for (int i = 1; i <= n; i++) idx[i] = i;
  sort(idx + 1, idx + n + 1, [](int x, int y) { return mx[x] > mx[y]; });
  priority_queue<pair<int, int> > pq;
  for (int i = k, j = 1; i >= 1; i--) {
    while (j <= n && mx[idx[j]] >= i)
      pq.push(make_pair(mi[idx[j]], idx[j])), j++;
    while (!pq.empty() && pq.top().first > i) pq.pop();
    if (pq.empty()) throw 1;
    r[pq.top().second] = i;
    pq.pop();
  }
}
int main() {
  int n, m;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &r[i]);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    e[x].push_back(y);
    f[y].push_back(x);
  }
  try {
    solve(n);
    remark(n);
    solve(n);
  } catch (...) {
    return 0 * printf("-1\n");
  }
  for (int i = 1; i <= n; i++) printf("%d%c", mi[i], " \n"[i == n]);
  return 0;
}