Problem: You are given a weighted directed graph, consisting of n vertices and m edges. You should answer q queries of two types:

  * 1 v — find the length of shortest path from vertex 1 to vertex v. 
  * 2 c l1 l2 ... lc — add 1 to weights of edges with indices l1, l2, ..., lc. 

Input

The first line of input data contains integers n, m, q (1 ≤ n, m ≤ 105, 1 ≤ q ≤ 2000) — the number of vertices and edges in the graph, and the number of requests correspondingly.

Next m lines of input data contain the descriptions of edges: i-th of them contains description of edge with index i — three integers ai, bi, ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 109) — the beginning and the end of edge, and its initial weight correspondingly.

Next q lines of input data contain the description of edges in the format described above (1 ≤ v ≤ n, 1 ≤ lj ≤ m). It's guaranteed that inside single query all lj are distinct. Also, it's guaranteed that a total number of edges in all requests of the second type does not exceed 106.

Output

For each query of first type print the length of the shortest path from 1 to v in a separate line. Print -1, if such path does not exists.

Examples

Input

3 2 9
1 2 0
2 3 0
2 1 2
1 3
1 2
2 1 1
1 3
1 2
2 2 1 2
1 3
1 2


Output

1
0
2
1
4
2


Input

5 4 9
2 3 1
2 4 1
3 4 1
1 2 0
1 5
1 4
2 1 2
2 1 2
1 4
2 2 1 3
1 4
2 1 4
1 4


Output

-1
1
2
3
4

Note

The description of changes of the graph in the first sample case:

<image>

The description of changes of the graph in the second sample case:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int MOD2 = 1007681537;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
const int maxn = 1e5 + 5;
int n, m, q;
vector<pair<int, int> > adj[maxn];
int w[maxn];
long long d[maxn];
int vis[maxn];
int ver[maxn];
vector<int> g[maxn];
void dijktra() {
  fill_n(d, n, LINF);
  d[0] = 0;
  priority_queue<pair<long long, int> > pq;
  pq.push(make_pair(-d[0], 0));
  while (int((pq).size())) {
    long long t = -pq.top().first;
    int u = pq.top().second;
    pq.pop();
    if (d[u] != t) continue;
    for (int i = (0); i < (int((adj[u]).size())); i++) {
      int v = adj[u][i].first;
      int ix = adj[u][i].second;
      if (chkmin(d[v], d[u] + w[ix])) {
        pq.push(make_pair(-d[v], v));
      }
    }
  }
}
int compare(pair<int, int> ii, pair<int, int> jj) {
  int i = ii.first, j = jj.first;
  if (d[i] != d[j]) return d[i] < d[j];
  return i < j;
}
int compare2(int i, int j) {
  if (d[i] != d[j]) return d[i] < d[j];
  return i < j;
}
void solve() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = (0); i < (m); i++) {
    int u, v;
    scanf("%d%d%d", &u, &v, w + i);
    u--, v--;
    adj[u].push_back(make_pair(v, i));
  }
  vector<int> qs;
  for (int i = (0); i < (q); i++) {
    int op;
    scanf("%d", &op);
    qs.push_back(op);
    if (op == 1) {
      scanf("%d", ver + i);
      ver[i]--;
    } else {
      int c;
      scanf("%d", &c);
      g[i].resize(c);
      for (int j = (0); j < (c); j++) {
        scanf("%d", &g[i][j]), g[i][j]--;
        w[g[i][j]]++;
      }
    }
  }
  vector<int> vs;
  for (int i = (0); i < (n); i++) vs.push_back(i);
  dijktra();
  vector<long long> res;
  int cnt = 0;
  for (int it = (q)-1; it >= (0); it--) {
    int op = qs[it];
    if (op == 1) {
      if (cnt % 50 == 0) {
        for (int u = (0); u < (n); u++)
          sort((adj[u]).begin(), (adj[u]).end(), compare);
        sort((vs).begin(), (vs).end(), compare2);
      }
      cnt++;
      int v = ver[it];
      static int que[10000000];
      int qh = 0, qe = 0;
      for (int i = (0); i < (n); i++) {
        int u = vs[i];
        if (d[u] < LINF) {
          vis[u] = 1, que[qe++] = u;
        }
      }
      while (qh < qe) {
        int u = que[qh++];
        for (int i = (0); i < (int((adj[u]).size())); i++) {
          int v = adj[u][i].first;
          int ix = adj[u][i].second;
          if (d[v] > d[u] + w[ix]) {
            d[v] = d[u] + w[ix];
            if (!vis[v]) {
              vis[v] = 1, que[qe++] = v;
            }
          }
        }
        vis[u] = 0;
      }
      if (d[v] == LINF) {
        res.push_back(-1);
      } else {
        res.push_back(d[v]);
      }
    } else {
      for (int i = (0); i < (int((g[it]).size())); i++) {
        int ix = g[it][i];
        w[ix]--;
      }
    }
  }
  reverse((res).begin(), (res).end());
  for (int i = (0); i < (int((res).size())); i++) printf("%I64d\n", res[i]);
}
int main() {
  int JUDGE_ONLINE = 1;
  if (fopen("in.txt", "r")) {
    JUDGE_ONLINE = 0;
    assert(freopen("in.txt", "r", stdin));
  } else {
  }
  solve();
  if (!JUDGE_ONLINE) {
  }
  return 0;
}