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 MXN = (int)2e5 + 7;
int n, m, k;
int rk[MXN];
bitset<MXN> in, out;
vector<int> edge[MXN], re_edge[MXN];
void dfs(int u) {
  in[u] = 1;
  for (int v : edge[u]) {
    if (!in[v])
      dfs(v);
    else if (!out[v]) {
      puts("-1");
      exit(0);
    }
  }
  out[u] = 1;
  return;
}
int rr[MXN], ll[MXN];
int deg[MXN];
void bye() {
  puts("-1");
  exit(0);
}
void build() {
  queue<int> Q;
  for (int i = 1; i < n + 1; i++) deg[i] = ((int)(re_edge[i]).size());
  for (int i = 1; i < n + 1; i++)
    if (!deg[i]) {
      Q.push(i);
      if (!rk[i]) rr[i] = k;
    }
  for (int i = 1; i < n + 1; i++)
    if (rk[i]) rr[i] = rk[i];
  while (!Q.empty()) {
    int v = Q.front();
    Q.pop();
    for (int u : edge[v]) {
      if (!(--deg[u])) Q.push(u);
      if (rk[u] && rr[v] - 1 < rk[u]) {
        puts("-1");
        exit(0);
      }
      if (!rr[u])
        rr[u] = rr[v] - 1;
      else
        rr[u] = min(rr[u], rr[v] - 1);
    }
  }
  for (int i = 1; i < n + 1; i++) deg[i] = ((int)(edge[i]).size());
  for (int i = 1; i < n + 1; i++)
    if (!deg[i]) {
      Q.push(i);
      if (!rk[i]) ll[i] = 1;
    }
  for (int i = 1; i < n + 1; i++)
    if (rk[i]) ll[i] = rk[i];
  while (!Q.empty()) {
    int v = Q.front();
    Q.pop();
    for (int u : re_edge[v]) {
      if (!(--deg[u])) Q.push(u);
      if (rk[u] && ll[v] + 1 > rk[u]) {
        puts("-1");
        exit(0);
      }
      ll[u] = max(ll[u], ll[v] + 1);
    }
  }
  for (int i = 1; i < n + 1; i++) {
    ll[i] = max(ll[i], 1);
    rr[i] = min(rr[i], k);
    if (ll[i] > rr[i]) bye();
  }
  return;
}
bool ck[MXN];
set<pair<pair<int, int>, int> > buf;
set<pair<int, int> > st;
queue<int> Q;
void update(int i) {
  while (!buf.empty()) {
    if (buf.begin()->first.first <= i) {
      pair<int, int> tmp =
          make_pair(buf.begin()->first.second, buf.begin()->second);
      st.insert(tmp);
      buf.erase(buf.begin());
    } else
      return;
  }
  return;
}
void update2() {
  while (!Q.empty()) {
    int v = Q.front();
    Q.pop();
    if (rk[v]) {
      for (int u : re_edge[v]) {
        if (!(--deg[u])) buf.insert(make_pair(make_pair(ll[u], rr[u]), u));
      }
    } else
      buf.insert(make_pair(make_pair(ll[v], rr[v]), v));
  }
  return;
}
void sol() {
  for (int i = 1; i < n + 1; i++) ck[rk[i]] = 1;
  for (int i = 1; i < n + 1; i++) deg[i] = ((int)(edge[i]).size());
  for (int i = 1; i < n + 1; i++)
    if (!deg[i]) Q.push(i);
  for (int i = 1; i < k + 1; i++) {
    bool ok = 0;
    update2();
    update(i);
    while (!st.empty()) {
      if (st.begin()->first < i) {
        int v = st.begin()->second;
        rk[v] = st.begin()->first;
        st.erase(st.begin());
        for (int u : re_edge[v]) {
          if (!(--deg[u])) Q.push(u);
          if (rk[u]) {
            if (rk[u] <= rk[v]) bye();
          } else {
            ll[u] = max(ll[u], rk[v] + 1);
            if (ll[u] > rr[u]) bye();
          }
        }
        update2();
        update(i);
      } else
        break;
    }
    if (ck[i]) continue;
    if (!st.empty()) {
      int v = st.begin()->second;
      rk[v] = i;
      st.erase(st.begin());
      ok = 1;
      for (int u : re_edge[v]) {
        if (!(--deg[u])) Q.push(u);
        if (rk[u]) {
          if (rk[u] <= rk[v]) bye();
        } else {
          ll[u] = max(ll[u], rk[v] + 1);
          if (ll[u] > rr[u]) bye();
        }
      }
      update2();
    }
    if (!ok) bye();
  }
  while (!st.empty()) {
    int v = st.begin()->second;
    rk[v] = st.begin()->first;
    st.erase(st.begin());
    for (int u : re_edge[v]) {
      if (!(--deg[u])) Q.push(u);
      if (rk[u]) {
        if (rk[u] <= rk[v]) bye();
      } else {
        ll[u] = max(ll[u], rk[v] + 1);
        if (ll[u] > rr[u]) bye();
      }
    }
    update2();
    update(k);
  }
  return;
}
int main(void) {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i < n + 1; i++) scanf("%d", rk + i);
  for (int i = 1; i < m + 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    edge[u].push_back(v);
    re_edge[v].push_back(u);
  }
  for (int i = 1; i < n + 1; i++)
    if (!in[i]) dfs(i);
  build();
  sol();
  for (int i = 1; i < n + 1; i++) printf("%d%c", rk[i], " \n"[i == n]);
  return 0;
}