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;
int a[200010];
struct EDGE {
  int to, next;
} e[200010];
int n, head[200010], top, deg[200010];
void add(int u, int v) {
  deg[v]++;
  e[top].to = v;
  e[top].next = head[u];
  head[u] = top++;
}
int b[200010];
queue<int> q;
void topo() {
  int i;
  for (i = 1; i <= n; i++) {
    if (!deg[i]) {
      q.push(i);
    }
  }
  int btop = 1, tmp;
  while (!q.empty()) {
    tmp = q.front();
    b[btop++] = tmp;
    q.pop();
    for (i = head[tmp]; ~i; i = e[i].next) {
      deg[e[i].to]--;
      if (!deg[e[i].to]) q.push(e[i].to);
    }
  }
  if (btop <= n) {
    cout << "-1" << endl;
    exit(0);
  }
}
int L[200010], R[200010], ans[200010];
struct _ {
  int id, l, r;
  bool operator<(const _ &ob) const { return r > ob.r; }
} c[200010];
bool cmp(const _ &a, const _ &b) { return a.l < b.l; }
priority_queue<_> pq;
int main() {
  memset(head, 255, sizeof(head));
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int m, k;
  cin >> n >> m >> k;
  int i;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int u, v;
  for (i = 1; i <= m; i++) {
    cin >> u >> v;
    add(u, v);
  }
  topo();
  int j;
  for (i = 1; i <= n; i++) {
    if (!a[i])
      L[i] = 1, R[i] = k;
    else
      L[i] = R[i] = a[i];
  }
  for (i = 1; i <= n; i++) {
    for (j = head[b[i]]; ~j; j = e[j].next) {
      R[e[j].to] = min(R[e[j].to], R[b[i]] - 1);
    }
  }
  for (i = n; i >= 1; i--) {
    for (j = head[b[i]]; ~j; j = e[j].next) {
      L[b[i]] = max(L[b[i]], L[e[j].to] + 1);
    }
  }
  for (i = 1; i <= n; i++) {
    if (L[i] > R[i]) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (i = 1; i <= n; i++) {
    c[i] = (_){i, L[i], R[i]};
  }
  sort(c + 1, c + 1 + n, cmp);
  int o = 1;
  for (i = 1; i <= k; i++) {
    while (o <= n) {
      if (c[o].l > i) break;
      pq.push(c[o]);
      o++;
    }
    bool flag = 0;
    _ tmp;
    while (!pq.empty()) {
      tmp = pq.top();
      if (tmp.r == i) {
        ans[tmp.id] = i;
        pq.pop();
        flag = 1;
        continue;
      } else {
        if (!flag) {
          flag = 1;
          ans[tmp.id] = i;
          pq.pop();
        }
        break;
      }
    }
    if (!flag) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
  cout << endl;
  return 0;
}