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;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= 1000000007;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
int gcd(int x, int y) {
  while (y) {
    int t = x % y;
    x = y;
    y = t;
  }
  return x;
}
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
namespace FastIO {
bool IOError = 0;
inline char NextChar() {
  static char buf[100000], *p1 = buf + 100000, *pend = buf + 100000;
  if (p1 == pend) {
    p1 = buf;
    pend = buf + fread(buf, 1, 100000, stdin);
    if (pend == p1) {
      IOError = 1;
      return -1;
    }
  }
  return *p1++;
}
inline bool Blank(char c) {
  return c == ' ' || c == '\n' || c == '\r' || c == '\t';
}
inline void read(int &x) {
  bool sign = 0;
  char ch = NextChar();
  x = 0;
  for (; Blank(ch); ch = NextChar())
    ;
  if (IOError) return;
  if (ch == '-') sign = 1, ch = NextChar();
  for (; ch >= '0' && ch <= '9'; ch = NextChar()) x = x * 10 + ch - '0';
  if (sign) x = -x;
}
inline void read(long long &x) {
  bool sign = 0;
  char ch = NextChar();
  x = 0;
  for (; Blank(ch); ch = NextChar())
    ;
  if (IOError) return;
  if (ch == '-') sign = 1, ch = NextChar();
  for (; ch >= '0' && ch <= '9'; ch = NextChar()) x = x * 10 + ch - '0';
  if (sign) x = -x;
}
inline void read(double &x) {
  bool sign = 0;
  char ch = NextChar();
  x = 0;
  for (; Blank(ch); ch = NextChar())
    ;
  if (IOError) return;
  if (ch == '-') sign = 1, ch = NextChar();
  for (; ch >= '0' && ch <= '9'; ch = NextChar()) x = x * 10 + ch - '0';
  if (ch == '.') {
    double tmp = 1;
    ch = NextChar();
    for (; ch >= '0' && ch <= '9'; ch = NextChar())
      tmp /= 10.0, x += tmp * (ch - '0');
  }
  if (sign) x = -x;
}
inline void read(char *s) {
  char ch = NextChar();
  for (; Blank(ch); ch = NextChar())
    ;
  if (IOError) return;
  for (; !Blank(ch) && !IOError; ch = NextChar()) *s++ = ch;
  *s = 0;
}
inline void read(char &c) {
  for (c = NextChar(); Blank(c); c = NextChar())
    ;
  if (IOError) {
    c = -1;
    return;
  }
}
}  // namespace FastIO
using namespace FastIO;
int q[601000], l[601000], r[601000], in[601000];
vector<int> g[601000], f[601000];
set<pair<int, int> > al;
int vis[601000], ans[601000];
int main() {
  int n, m, k;
  read(n), read(m), read(k);
  for (int i = 1; i <= n; i++) {
    int x;
    read(x);
    if (x)
      l[i] = r[i] = ans[i] = x;
    else
      l[i] = 1, r[i] = k;
  }
  for (int i = 0; i < m; i++) {
    int u, v;
    read(u), read(v);
    g[u].push_back(v);
    in[v]++;
  }
  int head = 0, tail = 1;
  for (int i = 1; i <= n; i++)
    if (!in[i]) q[tail++] = i;
  while (head < tail) {
    int u = q[++head];
    for (int v : g[u]) {
      in[v]--;
      if (!in[v]) q[tail++] = v;
    }
  }
  if (tail != n + 1) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int u = q[i];
    for (int v : g[u]) chkmin(r[v], r[u] - 1);
  }
  for (int i = n; i; i--) {
    int u = q[i];
    for (int v : g[u]) chkmax(l[u], l[v] + 1);
  }
  for (int i = 1; i <= n; i++) {
    f[l[i]].push_back(i);
    if (l[i] > r[i]) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j : f[i]) al.insert(pair<int, int>(r[j], j));
    if (al.empty()) {
      puts("-1");
      return 0;
    }
    set<pair<int, int> >::iterator it = al.begin();
    ans[(*it).second] = i;
    al.erase(it);
    while (!al.empty()) {
      set<pair<int, int> >::iterator it = al.begin();
      if ((*it).first > i) break;
      al.erase(it);
    }
  }
  for (int i = n; i; i--) {
    int u = q[i];
    int k = ans[u];
    for (int v : g[u]) {
      if (k && ans[v] >= ans[u]) {
        puts("-1");
        return 0;
      }
      if (!k) chkmax(ans[u], ans[v] + 1);
    }
    if (!ans[u]) ans[u] = 1;
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}