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 <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <map>
#include <list>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <sstream>
#include <fstream>
#include <iostream>
#include <algorithm>
using namespace std;

int *c;

struct job {
	int id;
	job(int i){
		this->id = i;
	}
	bool operator<(job j) const {
		return c[this->id] > c[j.id];
	}
};

int main() {
	int n;
	scanf("%d", &n);
	
	c = new int[n];
	vector<vector<pair<int, int> > > paths;
	
	for(int i = 0; i < n; i++){
		int u, k;
		scanf("%d %d", &u, &k);
		
		vector<pair<int, int> > path;
		c[u] = u ? INT_MAX : 0;
		while(k--){
			int v, c;
			scanf("%d %d", &v, &c);
			path.push_back(make_pair(v, c));
		}
		paths.push_back(path);
	}
	
	priority_queue<job> q;
	q.push(job(0));
	while(q.size()){
		int i = q.top().id;
		q.pop();
		
		for(int j = 0; j < paths[i].size(); j++){
			int dest = paths[i][j].first;
			int cost = paths[i][j].second;
			if(c[dest] > c[i] + cost){
				c[dest] = c[i] + cost;
				q.push(job(dest));
			}
		}
	}
	
	for(int i = 0; i < n; i++){
		printf("%d %d\n", i, c[i]);
	}
}