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 P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 10;
int n, m, k, a[N], deg1[N], deg2[N];
int L[N], R[N], vis[N];
int first[N], second[N], b[N];
vector<int> g1[N], g2[N];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    if (a[i])
      L[i] = R[i] = a[i];
    else
      L[i] = 1, R[i] = k;
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", first + i, second + i);
    g2[first[i]].push_back(second[i]), ++deg2[second[i]];
    g1[second[i]].push_back(first[i]), ++deg1[first[i]];
  }
  queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (!deg1[i]) q.push(i);
  int cnt = 0;
  while (q.size()) {
    int u = q.front();
    q.pop();
    if (L[u] > R[u]) return puts("-1"), 0;
    ++cnt;
    for (auto &v : g1[u]) {
      L[v] = max(L[v], L[u] + 1);
      if (!--deg1[v]) q.push(v);
    }
  }
  if (cnt != n) return puts("-1"), 0;
  for (int i = 1; i <= n; ++i)
    if (!deg2[i]) q.push(i);
  while (q.size()) {
    int u = q.front();
    q.pop();
    if (L[u] > R[u]) return puts("-1"), 0;
    for (auto &v : g2[u]) {
      R[v] = min(R[v], R[u] - 1);
      if (!--deg2[v]) q.push(v);
    }
  }
  for (int i = 1; i <= n; ++i) b[i] = i;
  sort(b + 1, b + 1 + n, [](int a, int b) { return L[a] < L[b]; });
  int now = 1;
  set<pair<int, int> > s;
  for (int i = 1; i <= k; ++i) {
    while (now <= n && L[b[now]] <= i) {
      s.insert(pair<int, int>(R[b[now]], b[now]));
      ++now;
    }
    int ok = 0;
    while (s.size()) {
      int id = s.begin()->second;
      if (ok && R[id] > i) break;
      ok = 1;
      s.erase(s.begin());
      a[id] = i;
    }
    if (!ok) return puts("-1"), 0;
  }
  for (int i = 1; i <= n; ++i) printf("%d ", a[i] ? a[i] : L[i]);
  putchar(10);
}