For a given weighted graph $G = (V, E)$, find the shortest path from a source to each vertex. For each vertex $u$, print the total weight of edges on the shortest path from vertex $0$ to $u$.

Constraints

* $1 \leq n \leq 10,000$
* $0 \leq c_i \leq 100,000$
* $|E| < 500,000$
* All vertices are reachable from vertex $0$

Input

In the first line, an integer $n$ denoting the number of vertices in $G$ is given. In the following $n$ lines, adjacency lists for each vertex $u$ are respectively given in the following format:

$u$ $k$ $v_1$ $c_1$ $v_2$ $c_2$ ... $v_k$ $c_k$

Vertices in $G$ are named with IDs $0, 1, ..., n-1$. $u$ is ID of the target vertex and $k$ denotes its degree. $v_i (i = 1, 2, ... k)$ denote IDs of vertices adjacent to $u$ and $c_i$ denotes the weight of a directed edge connecting $u$ and $v_i$ (from $u$ to $v_i$).

Output

For each vertex, print its ID and the distance separated by a space character in a line respectively. Print in order of vertex IDs.

Example

Input

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


Output

0 0
1 2
2 2
3 1
4 3
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF = 1000 * 100000;

int main(void){
  int n;
  cin >> n;
  vector< vector<pair<int, int> > > g(n);
  for(int i=0; i < n; i++){
    int v; cin >> v;
    int k; cin >> k;
    for(int j=0; j < k; j++){
      int v0, c; cin >> v0 >> c;
      g[v].push_back(make_pair(v0, c));
    }
  }
  vector<ll> d(n, INF);
  vector<bool> used(n, false);
  priority_queue<pair <int, int> > pq;
  d[0] = 0;

  pq.push(make_pair(0, 0));

  while(!pq.empty()){
    pair<int, int> f = pq.top(); pq.pop();
    int u = f.second;
    used[u] = true;
    if(d[u] < f.first * (-1)) continue;

    for(int i=0; i < g[u].size(); i++){
      int v = g[u][i].first;
      if(used[v] == true) continue;
      if(d[v] > d[u] + g[u][i].second){
        d[v] = d[u] + g[u][i].second;
        pq.push(make_pair(d[v] * (-1), v));
      }
    }
  }
  for(int i=0; i < n; i++){
    cout << i << " " << d[i] << endl;
  }

  return 0;
}