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 get() {
  char ch;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  if (ch == '-') {
    int s = 0;
    while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
    return -s;
  }
  int s = ch - '0';
  while (ch = getchar(), ch >= '0' && ch <= '9') s = s * 10 + ch - '0';
  return s;
}
const int N = 2e+5 + 5;
int n, m, k;
int rk[N];
struct edge {
  int x, nxt;
} e1[N], e2[N];
int h1[N], h2[N], tot1, tot2;
bool bz[N];
int Rig[N], Lef[N];
int que[N], r[N];
void inse1(int x, int y) {
  e1[++tot1].x = y;
  e1[tot1].nxt = h1[x];
  h1[x] = tot1;
}
void inse2(int x, int y) {
  e2[++tot2].x = y;
  e2[tot2].nxt = h2[x];
  h2[x] = tot2;
}
bool cmp(int x, int y) { return Rig[x] > Rig[y]; }
int main() {
  n = get();
  m = get();
  k = get();
  for (int i = 1; i <= n; i++) bz[rk[i] = get()] = 1;
  for (int i = 1; i <= m; i++) {
    int x = get(), y = get();
    inse1(x, y);
    inse2(y, x);
    r[y]++;
  }
  int he = 0, ta = 0;
  for (int i = 1; i <= n; i++) Rig[i] = k;
  for (int i = 1; i <= n; i++)
    if (!r[i]) que[++ta] = i;
  for (; he < ta;) {
    int x = que[++he];
    if (rk[x]) Rig[x] = min(Rig[x], rk[x]);
    for (int p = h1[x]; p; p = e1[p].nxt) {
      r[e1[p].x]--;
      Rig[e1[p].x] = min(Rig[e1[p].x], Rig[x] - 1);
      if (!r[e1[p].x]) que[++ta] = e1[p].x;
    }
  }
  if (ta != n) return printf("-1\n"), 0;
  sort(que + 1, que + 1 + n, cmp);
  set<int> s;
  s.clear();
  for (int i = 1; i <= k; i++)
    if (!bz[i]) s.insert(i);
  for (int i = n; i >= 1; i--) {
    int x = que[i];
    if (!rk[x]) {
      set<int>::iterator h = s.lower_bound(Lef[x]);
      if ((h != s.end()) && (*h) <= Rig[x]) {
        bz[rk[x] = (*h)] = 1;
        s.erase(h);
      } else
        rk[x] = max(1, Lef[x]);
    }
    if (rk[x] < Lef[x]) return printf("-1\n"), 0;
    if (rk[x] > Rig[x]) return printf("-1\n"), 0;
    for (int p = h2[x]; p; p = e2[p].nxt)
      Lef[e2[p].x] = max(Lef[e2[p].x], rk[x] + 1);
  }
  for (int i = 1; i <= k; i++)
    if (!bz[i]) return printf("-1\n"), 0;
  for (int i = 1; i <= n; i++) printf("%d ", rk[i]);
  return 0;
}