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];
int ww[maxn];
long long pot[maxn];
int d[maxn];
int vis[maxn];
void dijktra() {
  fill_n(pot + 1, n - 1, LINF);
  priority_queue<pair<long long, int> > pq;
  pq.push(make_pair(-pot[0], 0));
  while (int((pq).size())) {
    long long t = -pq.top().first;
    int u = pq.top().second;
    pq.pop();
    if (pot[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(pot[v], pot[u] + w[ix])) {
        pq.push(make_pair(-pot[v], v));
      }
    }
  }
}
void bfs(int k) {
  vector<queue<int> > que(k + 1);
  fill_n(d + 1, n - 1, INF);
  que[0].push(0);
  for (int i = (0); i < (k + 1); i++) {
    while (int((que[i]).size())) {
      int u = que[i].front();
      que[i].pop();
      if (d[u] != i) continue;
      for (int j = (0); j < (int((adj[u]).size())); j++) {
        int v = adj[u][j].first;
        int ix = adj[u][j].second;
        if (d[v] > d[u] + ww[ix] && d[u] + ww[ix] <= k) {
          d[v] = d[u] + ww[ix];
          que[d[v]].push(v);
        }
      }
    }
  }
}
void upd() {
  for (int u = (0); u < (n); u++)
    if (pot[u] < LINF) pot[u] += d[u];
  for (int u = (0); u < (n); u++)
    if (pot[u] < LINF) {
      for (int i = (0); i < (int((adj[u]).size())); i++) {
        int v = adj[u][i].first;
        int ix = adj[u][i].second;
        long long x = pot[u] + w[ix] - pot[v];
        ww[ix] = x > INF ? INF : x;
      }
    }
}
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));
  }
  dijktra(), upd();
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int u;
      scanf("%d", &u), u--;
      if (pot[u] == LINF) {
        puts("-1");
      } else {
        printf("%I64d\n", pot[u]);
      }
    } else {
      int k;
      scanf("%d", &k);
      for (int i = (0); i < (k); i++) {
        int ix;
        scanf("%d", &ix);
        ix--;
        w[ix]++, ww[ix]++;
      }
      bfs(k), upd();
    }
  }
}
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;
}