FF is a bad boy, he is always wooing TT to play the following game with him. This is a very humdrum game. To begin with, TT should write down a sequence of integers-_-!!(bored).

Then, FF can choose a continuous subsequence from it(for example the subsequence from the third to the fifth integer inclusively). After that, FF will ask TT what the sum of the subsequence he chose is. The next, TT will answer FF's question. Then, FF can redo this process. In the end, FF must work out the entire sequence of integers.

Boring~~Boring~~a very very boring game!!! TT doesn't want to play with FF at all. To punish FF, she often tells FF the wrong answers on purpose.

The bad boy is not a fool man. FF detects some answers are incompatible. Of course, these contradictions make it difficult to calculate the sequence.

However, TT is a nice and lovely girl. She doesn't have the heart to be hard on FF. To save time, she guarantees that the answers are all right if there is no logical mistakes indeed.

What's more, if FF finds an answer to be wrong, he will ignore it when judging next answers.

But there will be so many questions that poor FF can't make sure whether the current answer is right or wrong in a moment. So he decides to write a program to help him with this matter. The program will receive a series of questions from FF together with the answers FF has received from TT. The aim of this program is to find how many answers are wrong. Only by ignoring the wrong answers can FF work out the entire sequence of integers. Poor FF has no time to do this job. And now he is asking for your help~(Why asking trouble for himself~~Bad boy)

Line 2..M+1: Line i+1 contains three integer: Ai, Bi and Si. Means TT answered FF that the sum from Ai to Bi is Si. It's guaranteed that 0 < Ai <= Bi <= N.

You can assume that any sum of subsequence is fit in 32-bit integer.

---- sum[x]表示区间[x,f[x]]的和，这个可以在路径压缩的时候更新，对于一组数据（u,v,w),令r1=Find(u),r2=Find(v),于是若r1==r2,此时u,v就有了相同的参考点，而sum[u]为区间[u,r1(r2)]的和，sum[v]为区间[v,r2(r1)]的和，于是只需判断w==sum[v]-sum[u]即可;若r1<r2,此时可以分为两种情况，（u,v,r1,r2)或者(u,r1,v,r2),对于情况1来说，此时father[r1]=r2,sum[r1]=sum[v]-(sum[u]-w);对于情况2有sum[r1]=w-sun[u]+sum[v].通过观察，我们可以发现情况1和情况2的结果是一样的，于是可以合并。同理，对于r1>r2这种情况也一样，这里就不在赘述了。

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<algorithm>

 #define MAXN 222222

 using namespace std;

 int sum[MAXN],father[MAXN];

 int n,m;

 void Init(){

    memset(sum,,sizeof(sum));

   for(int i=;i<=n;i++)

     father[i]=i;

 }

 int Find(int x)

 {

     if(x==father[x])

         return x;

     int tmp=father[x];

     father[x]=Find(father[x]);

        sum[x]+=sum[tmp];

     return father[x];

 }

 bool Union(int u,int v,int w)

 {

     int r1=Find(u);

     int r2=Find(v);

     if(r1==r2){

         if(sum[u]==w+sum[v])

              return true;

         return false;

     }else {

         father[r1]=r2;

         sum[r1]=sum[v]-sum[u]+w;

         return true;

     }

 }

 int main()

 {

     int u,v,w,ans;

     while(~scanf("%d%d",&n,&m)){

         Init();

         ans=;

         while(m--){

             scanf("%d%d%d",&u,&v,&w);

             if(!Union(u-,v,w))ans++;

         }

         printf("%d\n",ans);

     }

     return ;

 }