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 maxn = 200005;
queue<int> q;
priority_queue<pair<int, int> > Q;
vector<pair<int, int> > V[maxn];
struct edge {
  int obj, last;
} e[maxn];
int n, N, m, p, top, cnt, point[maxn], in[maxn], ord[maxn], L[maxn], H[maxn],
    ans[maxn];
bool vis[maxn];
void add(int x, int y) {
  cnt++;
  e[cnt].obj = y;
  e[cnt].last = point[x];
  point[x] = cnt;
}
void Bye() {
  puts("-1");
  exit(0);
}
int main() {
  scanf("%d%d%d", &n, &m, &p);
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x);
    if (x)
      L[i] = x, H[i] = x;
    else
      L[i] = 1, H[i] = p;
  }
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%d%d", &x, &y);
    add(x, y);
    in[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (!in[i]) q.push(i), ord[++N] = i, vis[i] = true;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = point[u], v; i; i = e[i].last) {
      v = e[i].obj;
      in[v]--;
      if (!in[v]) q.push(v), ord[++N] = v, vis[v] = true;
    }
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) Bye();
  for (int i = 1, u; i <= n; i++) {
    u = ord[i];
    for (int j = point[u], v; j; j = e[j].last) {
      v = e[j].obj;
      H[v] = min(H[v], H[u] - 1);
    }
  }
  for (int i = n, u; i >= 1; i--) {
    u = ord[i];
    for (int j = point[u], v; j; j = e[j].last) {
      v = e[j].obj;
      L[u] = max(L[u], L[v] + 1);
    }
  }
  for (int i = 1; i <= n; i++)
    if (L[i] > H[i]) Bye();
  for (int i = 1; i <= n; i++) V[H[i]].push_back(pair<int, int>(L[i], i));
  top = p;
  while (top) {
    for (int i = 0, I = V[top].size(); i < I; i++) Q.push(V[top][i]);
    if (Q.empty()) Bye();
    ans[Q.top().second] = top;
    Q.pop();
    while (!Q.empty() && Q.top().first == top) {
      ans[Q.top().second] = top;
      Q.pop();
    }
    top--;
  }
  for (int i = 1; i < n; i++) printf("%d ", ans[i]);
  printf("%d\n", ans[n]);
  return 0;
}