As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input

Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (<= 500) – the number of cities (and the cities are numbered from 0 to N-1), M – the number of roads, C1 and C2 – the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.

Output

For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather.
All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input
5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1
Sample Output
2 4

dis[i]:起点到i的最短距离。

num[i]:从起点到i最短路径的条数。

totalWtg[i]:从起点到i的点权最大的最短路径的权重数。

``````
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <algorithm>
using namespace std;
int n, m, c1, c2;
int e, cityWtg, dis, num, totalWtg;
bool visit;
const int inf = 99999999;
int main() {
scanf("%d%d%d%d", &n, &m, &c1, &c2);
for (int i = 0; i < n; i++)
scanf("%d", &cityWtg[i]);
fill(e, e + 510 * 510, inf);
fill(dis, dis + 510, inf);
int a, b, c;
for (int i = 0; i < m; i++)
{
scanf("%d%d%d", &a, &b, &c);
e[a][b] = c;
e[b][a] = c;
}
dis[c1] = 0;
totalWtg[c1] = cityWtg[c1];
num[c1] = 1;
for (int i = 0; i < n; i++)
{
int u = -1, minn = inf;
for (int j = 0; j < n; j++)
{
if (visit[j] == false && dis[j] < minn) //没有visit过，且到起点的值最小，把u设为下一个current点
{
u = j;
minn = dis[j];
}
}
if (u == -1) break;
visit[u] = true;
for (int v = 0; v < n; v++)
{
if (visit[v] == false && e[u][v] != inf)
{
if (dis[u] + e[u][v] < dis[v])//当u的邻点+u到领点的距离小于之前路线保存的值
{ // 替换为新的最短路线
dis[v] = dis[u] + e[u][v];
num[v] = num[u];
totalWtg[v] = totalWtg[u] + cityWtg[v];
}
else if (dis[u] + e[u][v] == dis[v])
{
num[v] = num[v] + num[u];
if (totalWtg[u] + cityWtg[v] > totalWtg[v])// 如果点权大，取大点权
totalWtg[v] = totalWtg[u] + cityWtg[v];
}
}
}
}
printf("%d %d", num[c2], totalWtg[c2]);
return 0;
}
``````