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 = 2e+5 + 10;
const int INF = 0x7fffffff;
struct edge {
  int v, n;
} e[maxn * 2];
struct node {
  int mx, mi, i, c;
  bool operator<(const node &sec) const { return mx > sec.mx; }
} a[maxn];
int n, m, k;
int head[maxn], rd[maxn];
void read() {
  int i, u, v;
  scanf("%d%d%d", &n, &m, &k);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i].c);
    a[i].i = i;
    if (a[i].c)
      a[i].mx = a[i].mi = a[i].c;
    else
      a[i].mx = k, a[i].mi = 1;
  }
  memset(head, -1, sizeof(head));
  for (i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    e[i << 1].v = v;
    e[i << 1].n = head[u];
    head[u] = i << 1;
    e[i << 1 | 1].v = u;
    e[i << 1 | 1].n = head[v];
    head[v] = i << 1 | 1;
  }
}
void tuopu(int d) {
  int i, u, v;
  queue<int> q;
  memset(rd, 0, sizeof(rd));
  for (i = 0; i < m; i++) rd[e[i << 1 | d].v]++;
  for (i = 1; i <= n; i++)
    if (!rd[i]) q.push(i);
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (i = head[u]; i != -1; i = e[i].n)
      if ((i & 1) == d) {
        v = e[i].v;
        rd[v]--;
        if (!rd[v]) q.push(v);
        if (d)
          a[v].mi = max(a[v].mi, a[u].mi + 1);
        else
          a[v].mx = min(a[v].mx, a[u].mx - 1);
      }
  }
  for (i = 1; i <= n; i++)
    if (rd[i]) {
      printf("-1\n");
      exit(0);
    }
}
bool cmpa(node aa, node bb) { return aa.mi < bb.mi; }
bool cmpi(node aa, node bb) { return aa.i < bb.i; }
void check() {
  for (int i = 1; i <= n; i++)
    if (a[i].mx < a[i].mi) {
      printf("-1\n");
      exit(0);
    }
}
void solve() {
  int i, j = 1;
  priority_queue<node> q;
  bool del;
  for (i = 1; i <= k; i++) {
    del = false;
    for (; j <= n && a[j].mi == i; j++) {
      node tmp = a[j];
      tmp.i = j;
      q.push(tmp);
    }
    while (!q.empty() && q.top().mx == i) {
      node tmp = q.top();
      q.pop();
      a[tmp.i].c = i;
      del = true;
    }
    if (!del) {
      if (q.empty()) {
        printf("-1\n");
        exit(0);
      }
      node tmp = q.top();
      q.pop();
      a[tmp.i].c = i;
    }
  }
}
void print() {
  for (int i = 1; i <= n; i++) printf("%d ", a[i].c);
}
int main() {
  read();
  tuopu(0);
  tuopu(1);
  sort(a + 1, a + n + 1, cmpa);
  check();
  solve();
  sort(a + 1, a + n + 1, cmpi);
  print();
  return 0;
}