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 N = 200200;
struct Tsort {
  int l, t, i;
  bool operator<(const Tsort &t) const { return l < t.l; }
} sor[N];
struct Tedge {
  int v, nxt;
} e[N], e2[N];
int head[N], head2[N], low[N], top[N], key[N], ru[N], ru2[N], ans[N], cnt = 1,
                                                                      n, m, k;
bool fuck[N];
queue<int> q;
void add(int u, int v) {
  e[cnt].v = v;
  e[cnt].nxt = head[u];
  head[u] = cnt++;
}
void add2(int u, int v) {
  e2[cnt].v = v;
  e2[cnt].nxt = head2[u];
  head2[u] = cnt++;
}
void init() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &key[i]);
    low[i] = top[i] = key[i];
    if (key[i] == 0) low[i] = 1, top[i] = k;
  }
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(v, u);
    add2(u, v);
    ru[u]++;
    ru2[v]++;
  }
}
bool cmp(Tsort a, Tsort b) {
  if (a.t == b.t) return a.l > b.l;
  return a.t > b.t;
}
void solve() {
  for (int i = 1; i <= n; i++)
    if (!ru[i]) q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head[u]; i; i = e[i].nxt) {
      int v = e[i].v;
      low[v] = max(low[v], low[u] + 1);
      ru[v]--;
      if (!ru[v]) q.push(v);
    }
  }
  for (int i = 1; i <= n; i++)
    if (ru[i]) {
      printf("-1\n");
      return;
    }
  for (int i = 1; i <= n; i++)
    if (!ru2[i]) q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = head2[u]; i; i = e2[i].nxt) {
      int v = e2[i].v;
      top[v] = min(top[v], top[u] - 1);
      ru2[v]--;
      if (!ru2[v]) q.push(v);
    }
  }
  for (int i = 1; i <= n; i++)
    if (ru2[i]) {
      printf("-1\n");
      return;
    }
  for (int i = 1; i <= n; i++)
    if (low[i] > top[i]) {
      printf("-1\n");
      return;
    } else {
      sor[i].l = low[i];
      sor[i].t = top[i];
      sor[i].i = i;
    }
  sort(sor + 1, sor + 1 + n, cmp);
  priority_queue<Tsort> qq;
  for (int i = k, pos = 1; i > 0; i--) {
    bool flag = 0;
    while (sor[pos].t >= i) qq.push(sor[pos++]);
    while ((!qq.empty()) && qq.top().l == i) {
      flag = 1;
      ans[qq.top().i] = i, qq.pop();
    }
    if (!flag) {
      if (!qq.empty())
        ans[qq.top().i] = i, qq.pop();
      else {
        printf("-1\n");
        return;
      }
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  printf("\n");
}
int main() {
  init();
  solve();
  return 0;
}