#include <iostream>
#include <fstream>
#include <vector>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <cmath>
#include <cstring>
#include <ctime>
#include <unordered_map>
#include <iomanip>
#include <complex>
#include <cassert>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define deb(a) cerr<< #a << "= " << (a)<<"\n";

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef long double ld;
typedef complex<double> base;
typedef vector<int> vi;
typedef pair<int,int> pii;

template<class T> ostream& operator<<(ostream& stream, const vector<T> v){ stream << "[ "; for (int i=0; i<(int)v.size(); i++) stream << v[i] << " "; stream << "]"; return stream; }
ll fpow(ll x, ll p, ll m){ll r=1; for (;p;p>>=1){ if (p&1) r=r*x%m; x=x*x%m; } return r;}
ll inv(ll a, ll b){ return a<2 ? a : ((a-inv(b%a,a))*b+1)/a%b; }
int gcd(int a, int b){ if (!b) return a; return gcd(b,a%b);}
ll gcd(ll a, ll b){ if (!b) return a; return gcd(b,a%b);}


int N,M,S,D,C[1010][1010],F[1010][1010],d[1010],ptr[1010];
vi g[1010];

bool bfs(){
	vi q;
	memset(d,0,sizeof(d));
	q.pb(S);
	d[S]=1;
	
	for (int i=0; i<q.size(); i++){
		int nod=q[i];
		for (int nxt : g[nod])
			if (C[nod][nxt]-F[nod][nxt]>0 && !d[nxt]){
				d[nxt]=d[nod]+1;
				q.pb(nxt);
			}
	}
	
	return d[D]!=0;
}

int dfs(int nod, int val){
	if (val==0) return 0;
	if (nod==D) return val;
	
	for (int &i=ptr[nod]; i<g[nod].size(); i++){
		int nxt=g[nod][i];
		
		if (C[nod][nxt]-F[nod][nxt]>0 && d[nxt]==d[nod]+1){
			int nval=dfs(nxt,min(val,C[nod][nxt]-F[nod][nxt]));
			if (nval>0){
				F[nod][nxt]+=nval;
				F[nxt][nod]-=nval;
				return nval;
			}
		}
	}
	
	return 0;
}

int dinic(){
	int flow=0;
	
	while (bfs()){
		
		memset(ptr,0,sizeof(ptr));
		while (int v=dfs(S,(1<<30))){
			
			flow+=v;
		}
	}
	
	return flow;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	ifstream fin("maxflow.in");
	ofstream fout("maxflow.out");
	fin >> N >> M;
	
	S=1,D=N;
	
	for (int i=1; i<=M; i++){
		int x,y,c;
		fin >> x >> y >> c;
		g[x].pb(y);
		g[y].pb(x);
		
		C[x][y]=c;
	}
	
	fout << dinic() << "\n";
	
	return 0;
}