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 oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 1000 * 1000 * 1000 + 7;
const int N = 200105;
const double PI = acos(-1.0);
const double eps = 1e-6;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void file_put() {
  freopen("filename.in", "r", stdin);
  freopen("filename.out", "w", stdout);
}
struct ttype {
  int x, v;
};
inline bool operator<(const ttype &a, const ttype &b) { return a.v > b.v; }
int n, k, m, r[N], in[N], out[N], head[N], nxt[N], e[N],
    tot = 0, a[N], cnt = 0, ans[N], kk, x, y, _head[N], _nxt[N], _e[N],
    _tot = 0, up[N], down[N];
queue<int> Q;
priority_queue<ttype> QQ;
vector<int> V[N];
void I(int x, int y) { e[++tot] = y, nxt[tot] = head[x], head[x] = tot; }
void _I(int x, int y) {
  _e[++_tot] = y, _nxt[_tot] = _head[x], _head[x] = _tot;
}
bool topsort() {
  for (int i = (1); i <= (n); ++i)
    if (!in[i]) Q.push(i);
  while (!Q.empty()) {
    int x = Q.front();
    a[++cnt] = x;
    Q.pop();
    for (int i = head[x]; i; i = nxt[i]) {
      in[e[i]]--;
      if (!in[e[i]]) Q.push(e[i]);
    }
  }
  return cnt == n;
}
void color(int x, int k) {
  ans[x] = k;
  for (int i = _head[x]; i; i = _nxt[i]) {
    down[_e[i]] = ((down[_e[i]]) > (k + 1) ? (down[_e[i]]) : (k + 1));
    out[_e[i]]--;
    if (!out[_e[i]]) V[down[_e[i]]].push_back(_e[i]);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &kk);
  for (int i = (1); i <= (n); ++i)
    scanf("%d", &r[i]), up[i] = (r[i] ? r[i] : kk),
                        down[i] = ((r[i]) > (1) ? (r[i]) : (1));
  for (int i = (1); i <= (m); ++i)
    scanf("%d%d", &x, &y), in[y]++, out[x]++, I(x, y), _I(y, x);
  if (!topsort()) return 0 * printf("-1\n");
  for (int k = (1); k <= (n); ++k)
    for (int i = head[a[k]]; i; i = nxt[i])
      up[e[i]] = ((up[e[i]]) < (up[a[k]] - 1) ? (up[e[i]]) : (up[a[k]] - 1));
  for (int i = (1); i <= (n); ++i)
    if (up[i] <= 0 || r[i] && up[i] < r[i]) return 0 * printf("-1\n");
  for (int i = (1); i <= (n); ++i)
    if (!out[i]) V[down[i]].push_back(i);
  for (int i = (1); i <= (kk); ++i) {
    for (__typeof((V[i]).begin()) it = (V[i]).begin(); it != (V[i]).end(); it++)
      QQ.push(ttype{*it, up[*it]});
    if (QQ.empty()) return 0 * printf("-1\n");
    color(QQ.top().x, i), QQ.pop();
    while (!QQ.empty() && QQ.top().v == i) color(QQ.top().x, i), QQ.pop();
  }
  for (int i = (1); i <= (n - 1); ++i) printf("%d ", ans[i]);
  printf("%d\n", ans[n]);
  return 0;
}