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 = 200100;
struct edge {
  int obj;
  edge *Next;
} e[maxn * 3];
edge *head[maxn];
edge *nhead[maxn];
int cur = -1;
int Heap[maxn];
int tail = 0;
int Min[maxn];
int Max[maxn];
int ans[maxn];
int pin[maxn];
int pout[maxn];
int vis[maxn];
int que[maxn];
int he, ta;
int r[maxn];
int n, m, k;
void Add(edge **Head, int x, int y) {
  cur++;
  e[cur].obj = y;
  e[cur].Next = Head[x];
  Head[x] = e + cur;
}
void Dfs(int node) {
  vis[node] = 1;
  for (edge *p = head[node]; p; p = p->Next) {
    int to = p->obj;
    if (vis[to] == 1) {
      printf("-1\n");
      exit(0);
    }
    if (!vis[to]) Dfs(to);
  }
  vis[node] = 2;
}
void Calc_Max() {
  he = 0, ta = 1;
  for (int i = 1; i <= n; i++) {
    Max[i] = k;
    if (!pout[i]) que[++ta] = i;
    if (r[i]) Max[i] = r[i];
  }
  while (he < ta) {
    int node = que[++he];
    for (edge *p = nhead[node]; p; p = p->Next) {
      int to = p->obj;
      Max[to] = min(Max[to], Max[node] - 1);
      pout[to]--;
      if (!pout[to]) {
        que[++ta] = to;
        if (r[to] > Max[to] && r[to]) {
          printf("-1\n");
          exit(0);
        }
        if (r[to]) Max[to] = r[to];
      }
    }
  }
}
void Insert(int x) {
  Heap[++tail] = x;
  x = tail;
  while (x > 1) {
    int y = x >> 1;
    if (Max[Heap[y]] < Max[Heap[x]]) break;
    swap(Heap[x], Heap[y]);
    x = y;
  }
}
int Delete() {
  int temp = Heap[1];
  Heap[1] = Heap[tail];
  tail--;
  int x = 1;
  while (1) {
    int y = x, Left = x << 1, Right = Left | 1;
    if (Left <= tail && Max[Heap[Left]] < Max[Heap[y]]) y = Left;
    if (Right <= tail && Max[Heap[Right]] < Max[Heap[y]]) y = Right;
    if (x == y) break;
    swap(Heap[x], Heap[y]);
    x = y;
  }
  return temp;
}
void Update(int node) {
  for (edge *p = head[node]; p; p = p->Next) {
    int to = p->obj;
    pin[to]--;
    Min[to] = max(Min[to], ans[node] + 1);
    if (!pin[to]) {
      if (r[to] < Min[to] && r[to]) {
        printf("-1\n");
        exit(0);
      }
      if (r[to]) Min[to] = r[to];
      Add(nhead, Min[to], to);
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &r[i]), head[i] = nhead[i] = NULL;
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    Add(head, y, x);
    pin[x]++;
    pout[y]++;
    Add(nhead, x, y);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) Dfs(i);
  Calc_Max();
  for (int i = 1; i <= n; i++)
    if (Max[i] < 1) {
      printf("-1\n");
      return 0;
    }
  for (int i = 1; i <= k; i++) nhead[i] = NULL;
  for (int i = 1; i <= n; i++)
    if (!pin[i]) {
      Min[i] = 1;
      if (r[i]) Min[i] = r[i];
      Add(nhead, Min[i], i);
    }
  for (int i = 1; i <= k; i++) {
    for (edge *p = nhead[i]; p; p = p->Next) Insert(p->obj);
    if (!tail) {
      printf("-1\n");
      return 0;
    }
    int x = Delete();
    ans[x] = i;
    Update(x);
    while (tail && Max[Heap[1]] == i) x = Delete(), ans[x] = i, Update(x);
  }
  for (int i = 1; i <= n; i++)
    if (!ans[i]) {
      printf("-1\n");
      return 0;
    }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}