Jzzhu is the president of country A. There are n cities numbered from 1 to n in his country. City 1 is the capital of A. Also there are m roads connecting the cities. One can go from city ui to vi (and vise versa) using the i-th road, the length of this road is xi. Finally, there are k train routes in the country. One can use the i-th train route to go from capital of the country to city si (and vise versa), the length of this route is yi.

Jzzhu doesn't want to waste the money of the country, so he is going to close some of the train routes. Please tell Jzzhu the maximum number of the train routes which can be closed under the following condition: the length of the shortest path from every city to the capital mustn't change.

Input

The first line contains three integers n, m, k (2 ≤ n ≤ 105; 1 ≤ m ≤ 3·105; 1 ≤ k ≤ 105).

Each of the next m lines contains three integers ui, vi, xi (1 ≤ ui, vi ≤ n; ui ≠ vi; 1 ≤ xi ≤ 109).

Each of the next k lines contains two integers si and yi (2 ≤ si ≤ n; 1 ≤ yi ≤ 109).

It is guaranteed that there is at least one way from every city to the capital. Note, that there can be multiple roads between two cities. Also, there can be multiple routes going to the same city from the capital.

Output

Output a single integer representing the maximum number of the train routes which can be closed.

Examples

input

Copy

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

output

Copy

2

input

Copy

2 2 3 1 2 2 2 1 3 2 1 2 2 2 3

output

Copy

2 一次spfa 考虑对每个点，每次被更新时，有两种情况，铁路和公路，被铁路更新标记为1（被认为不能删），被公路更新就标记回0. 另一方面，当dis[v]==dis[u]+w时，如果已被标记为1，那还需要重新标记

#include 
        
         using namespace std;typedef long long ll;#define inf 2147483647const ll INF = 0x3f3f3f3f3f3f3f3fll;#define ri register inttemplate 
         
           inline T min(T a, T b, T c){    return min(min(a, b), c);}template 
          
            inline T max(T a, T b, T c){    return max(max(a, b), c);}template 
           
             inline T min(T a, T b, T c, T d){    return min(min(a, b), min(c, d));}template 
            
              inline T max(T a, T b, T c, T d){    return max(max(a, b), max(c, d));}#define scanf1(x) scanf("%d", &x)#define scanf2(x, y) scanf("%d%d", &x, &y)#define scanf3(x, y, z) scanf("%d%d%d", &x, &y, &z)#define scanf4(x, y, z, X) scanf("%d%d%d%d", &x, &y, &z, &X)#define pi acos(-1)#define me(x, y) memset(x, y, sizeof(x));#define For(i, a, b) for (int i = a; i <= b; i++)#define FFor(i, a, b) for (int i = a; i >= b; i--)#define bug printf("***********\n");#define mp make_pair#define pb push_backconst int N =1e6;const int M=200005;// name*******************************struct edge{    int to,nxt;    ll w;} e[N];int tot=1;int fst[N];ll dis[N];int ins[N];int vis[N];int pos[N];priority_queue
             
              que;int sum=0;int n,m,k;// function******************************void add(int u,int v,int w){ e[++tot].to=v; e[tot].nxt=fst[u]; fst[u]=tot; e[tot].w=w;}void spfa(){ For(i,1,n)dis[i]=INF; que.push(1); ins[1]=1; dis[1]=0; while(!que.empty()) { int u=que.top(); que.pop(); ins[u]=0; for(int p=fst[u]; p; p=e[p].nxt) { int v=e[p].to; ll w=e[p].w; if(dis[v]==dis[u]+w&&pos[v]) pos[v]=vis[p]; if(dis[v]>dis[u]+w) { dis[v]=dis[u]+w; pos[v]=vis[p]; if(!ins[v]) { ins[v]=1; que.push(v); } } } }}//***************************************int main(){// ios::sync_with_stdio(0);// cin.tie(0); // freopen("test.txt", "r", stdin); // freopen("outout.txt","w",stdout); scanf("%d%d%d",&n,&m,&k); For(i,1,m) { int u,v,w; scanf("%d%d%d",&u,&v,&w); add(u,v,w); add(v,u,w); } For(i,1,k) { int x,y; scanf("%d%d",&x,&y); add(1,x,y); vis[tot]=1; } spfa(); For(i,1,n) sum+=pos[i]; cout<