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;
template <class T>
inline void input(T &x) {
  register char c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
inline long long bigmod(long long p, long long e, long long M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
const int N = 2e5 + 5;
bool cycle = 0;
vector<int> top;
int vis[N];
int n, m, k;
vector<int> g[N];
int r[N];
vector<pair<int, int> > vec;
int high[N];
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int in[N];
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) cin >> r[i];
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    in[b]++;
  }
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (in[i] == 0) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    top.push_back(u);
    for (auto v : g[u]) {
      in[v]--;
      if (in[v] == 0) q.push(v);
    }
  }
  if (top.size() != n) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (r[i])
      high[i] = r[i];
    else
      high[i] = k;
  }
  for (auto v : top) {
    if (high[v] < 1) {
      printf("-1\n");
      return 0;
    }
    if (r[v]) {
      if (r[v] > high[v]) {
        printf("-1\n");
        return 0;
      }
    } else {
      vec.push_back({v, high[v]});
    }
    for (auto x : g[v]) {
      high[x] = min(high[x], high[v] - 1);
    }
  }
  set<int> second;
  for (int i = 1; i <= k; i++) second.insert(i);
  for (int i = 1; i <= n; i++)
    if (second.count(r[i])) second.erase(r[i]);
  sort((vec).begin(), (vec).end(), cmp);
  for (auto v : vec) {
    int now = v.first;
    int mx = v.second;
    int iMin = 1;
    for (auto x : g[now]) {
      if (r[x]) iMin = max(iMin, 1 + r[x]);
    }
    if (second.size()) {
      auto v = second.lower_bound(iMin);
      if (v != second.end() && *v <= mx) {
        r[now] = *v;
        second.erase(*v);
      } else
        r[now] = mx;
    } else
      r[now] = mx;
  }
  if (second.size() == 0) {
    for (int i = 1; i <= n; i++) printf("%d ", r[i]);
  } else
    printf("-1\n");
  return 0;
}