#pragma GCC optimize("Ofast,unroll-loops")
#include <bits/stdc++.h>

#define sz(a) ((int)(a).size())
#define all(a) (a).begin(), (a).end()

#define fi first
#define se second

using namespace std;

class InParser {
private:
	FILE *fin;
	char *buff;
	int sp;

	char read_ch() {
		++sp;
		if (sp == 4096) {
			sp = 0;
			fread(buff, 1, 4096, fin);
		}
		return buff[sp];
	}

public:
	InParser(const char* nume) {
		fin = fopen(nume, "r");
		buff = new char[4096]();
		sp = 4095;
	}

	InParser& operator >> (int &n) {
		char c;
		while (!isdigit(c = read_ch()) && c != '-');
		int sgn = 1;
		if (c == '-') {
			n = 0;
			sgn = -1;
		} else {
			n = c - '0';
		}
		while (isdigit(c = read_ch())) {
			n = 10 * n + c - '0';
		}
		n *= sgn;
		return *this;
	}

	InParser& operator >> (long long &n) {
		char c;
		n = 0;
		while (!isdigit(c = read_ch()) && c != '-');
		long long sgn = 1;
		if (c == '-') {
			n = 0;
			sgn = -1;
		} else {
			n = c - '0';
		}
		while (isdigit(c = read_ch())) {
			n = 10 * n + c - '0';
		}
		n *= sgn;
		return *this;
	}
};

#ifdef LOCAL
InParser fin("input.txt");
#define fout cout
#else
#define FILE_NAME "maxflow"
InParser fin(FILE_NAME ".in");
ofstream fout(FILE_NAME ".out");
#define endl '\n'
#endif

typedef long long ll;
typedef pair<int, int> pii;

const int NMAX = 1005;
const int INF = 2e9+5;

int n, m;
int g[NMAX][NMAX];
bool viz[NMAX];
int nxt[NMAX];

void read() {
	fin >> n >> m;
	for (int i = 1, x, y, c; i <= m; i++) {
		fin >> x >> y >> c;
		g[x][y] = c;
		g[y][x] = 0;
	}
}

int dfs(int x, int mn) {
	if (viz[x]) return INF;
	viz[x] = true;
	if (x == n) return mn;
	for (int i = 1; i <= n; i++) {
		int ret;
		if (g[x][i] && (ret = dfs(i, min(mn, g[x][i]))) != INF) {
			nxt[x] = i;
			return ret;
		}
	}
	return INF;
}

int solve() {
	int max_flow = 0, flow;
	while ((flow = dfs(1, INF)) != INF) {
		memset(viz, 0, n + 1);
		for (int v = 1; v != n; v = nxt[v]) {
			g[v][nxt[v]] -= flow;
			g[nxt[v]][v] += flow;
		}
		max_flow += flow;
	}
	return max_flow;
}

signed main() {
	read();
	fout << solve() << endl;
	return 0;
}