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;
int n, m, k, tot, head[200010], mx[200010], mn[200010], out[200010], in[200010],
    F[200010];
int c[200010], cnt = 1, b[200010], first, last;
struct rd {
  int next, to, f;
} road[400010];
struct node {
  int x, y, id;
  bool operator<(const node &a) const { return a.y < y; }
} a[200010];
inline bool cmp(node a, node b) { return a.x < b.x; }
priority_queue<node> q;
bool flag[200010];
inline void addedge(int x, int y, int z) {
  road[++tot].next = head[x];
  road[tot].to = y;
  road[tot].f = z;
  head[x] = tot;
}
inline void work1() {
  first = 1, last = 0;
  for (int i = 1; i <= n; i++)
    if (in[i] == 0) F[++last] = i;
  while (first <= last) {
    int x = F[first];
    flag[x] = 1;
    for (int now = head[x]; now; now = road[now].next)
      if (road[now].f == 1) {
        mx[road[now].to] = min(mx[road[now].to], mx[x] - 1);
        if (--in[road[now].to] == 0) F[++last] = road[now].to;
      }
    first++;
  }
}
inline void work2() {
  first = 1, last = 0;
  for (int i = 1; i <= n; i++)
    if (out[i] == 0) F[++last] = i;
  while (first <= last) {
    int x = F[first];
    for (int now = head[x]; now; now = road[now].next)
      if (road[now].f == -1) {
        mn[road[now].to] = max(mn[road[now].to], mn[x] + 1);
        if (--out[road[now].to] == 0) F[++last] = road[now].to;
      }
    first++;
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%d%d", &x, &y);
    addedge(x, y, 1);
    addedge(y, x, -1);
    out[x]++;
    in[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (c[i] == 0)
      mn[i] = 1, mx[i] = k;
    else
      mn[i] = mx[i] = c[i];
  work1();
  for (int i = 1; i <= n; i++)
    if (flag[i] == 0 || mn[i] > mx[i]) {
      printf("-1\n");
      return 0;
    }
  work2();
  for (int i = 1; i <= n; i++) a[i] = (node){mn[i], mx[i], i};
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1, j = 1; i <= k; i++) {
    while (j <= n && a[j].x <= i) q.push(a[j]), j++;
    int kill = 0;
    while (!q.empty()) {
      if (q.top().y == i || kill == 0)
        b[q.top().id] = i, kill = 1, q.pop();
      else
        break;
    }
    if (kill == 0) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = last; i >= 1; i--) {
    int x = F[i];
    if (b[x] > mx[x] || b[x] < mn[x]) {
      printf("-1\n");
      return 0;
    }
    for (int now = head[x]; now; now = road[now].next)
      if (road[now].f == 1) b[road[now].to] = min(b[road[now].to], b[x] - 1);
  }
  for (int i = 1; i <= n; i++) printf("%d ", b[i]);
  return 0;
}