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<iostream>
#include<algorithm>
#include<queue>
using namespace std;
static const int R =10000;
static const int IF = (1<<20);
static const int WH = 0;
static const int GR =1;
static const int BL =2;

int num;
vector<pair<int,int> > ax[R];

void dx(){
  priority_queue<pair<int,int> > Pi;
  int IR[R],i,j,y,x;
  int D[R];

  for(i=0;i<num;i++){
    D[i] = IF;
    IR[i] = WH;
  }

  D[0] = 0;
  Pi.push(make_pair(0,0));
  IR[0] =GR;

  while(!Pi.empty()){
    pair<int,int> F=Pi.top();
    Pi.pop();
    x =F.second;

    IR[x] =BL;

    if(D[x] < F.first * (-1)){
      continue;
    }

    for(j=0;j<ax[x].size();j++){
       y= ax[x][j].first;
       
      if(IR[y] ==BL){
	continue;
      }
      if(D[y] > D[x] + ax[x][j].second){
	 D[y] = D[x] + ax[x][j].second;
	 Pi.push(make_pair(D[y] * (-1),y));
	 IR[y] = GR;
	 }
    }
  }

  for(int i=0;i<num;i++){
    cout << i << " " << (D[i] == IF ? -1 : D[i] ) << endl;
  }
}

int main(){
  int z,x,y,ci,i,j;

  cin >> num;
  for(i=0;i<num;i++){
    cin >> x >> z;

    for(j=0;j<z;j++){
      cin >> y >> ci;
      ax[x].push_back(make_pair(y,ci));
    }
  }
  dx();

  return 0;
}

