Problem:
Score : 600 points

Problem Statement

Kenkoooo found a simple connected graph.
The vertices are numbered 1 through n.
The i-th edge connects Vertex u_i and v_i, and has a fixed integer s_i.

Kenkoooo is trying to write a positive integer in each vertex so that the following condition is satisfied:

For every edge i, the sum of the positive integers written in Vertex u_i and v_i is equal to s_i.

Find the number of such ways to write positive integers in the vertices.

Constraints

2 \leq n \leq 10^5

1 \leq m \leq 10^5

1 \leq u_i < v_i \leq n

2 \leq s_i \leq 10^9

If i\neq j, then u_i \neq u_j or v_i \neq v_j.

The graph is connected.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

n m
u_1 v_1 s_1
:
u_m v_m s_m

Output

Print the number of ways to write positive integers in the vertices so that the condition is satisfied.

Sample Input 1

3 3
1 2 3
2 3 5
1 3 4

Sample Output 1

1

The condition will be satisfied if we write 1,2 and 3 in vertices 1,2 and 3, respectively.
There is no other way to satisfy the condition, so the answer is 1.

Sample Input 2

4 3
1 2 6
2 3 7
3 4 5

Sample Output 2

3

Let a,b,c and d be the numbers to write in vertices 1,2,3 and 4, respectively.
There are three quadruples (a,b,c,d) that satisfy the condition:

(a,b,c,d)=(1,5,2,3)

(a,b,c,d)=(2,4,3,2)

(a,b,c,d)=(3,3,4,1)

Sample Input 3

8 7
1 2 1000000000
2 3 2
3 4 1000000000
4 5 2
5 6 1000000000
6 7 2
7 8 1000000000

Sample Output 3

0

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <array>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <algorithm>
#include <math.h>
#include <string>
#include <climits>
#include <assert.h>
#include <iomanip>
#include <bitset>
#include <queue>
#include <deque>
#include <stack>
#include <functional>
#include <fstream>
#include <random>

#define LEN(x) (long long)(x.size())
#define FOR(i, a, n) for(int i=(a);i<(n); ++i)
#define FOE(i, a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v, x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT_COUNT(bit) (__builtin_popcount(bit))

typedef long long LL;
template<typename T> std::vector<T> make_v(size_t a){return std::vector<T>(a);}
template<typename T,typename... Ts> auto make_v(size_t a, Ts... ts){ return std::vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));}    // C++14
template<typename T,typename V> typename std::enable_if<std::is_class<T>::value==0>::type fill_v(T &t,const V &v){t=v;}
template<typename T,typename V> typename std::enable_if<std::is_class<T>::value!=0>::type fill_v(T &t,const V &v){for(auto &e:t) fill_v(e,v);}
template<class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
void print() { std::cout << std::endl; }
template <class Head, class... Tail> void print(Head&& head, Tail&&... tail) { std::cout << head; if (sizeof...(tail) != 0) {std::cout << " ";} print(std::forward<Tail>(tail)...); }
template <class T> void print(std::vector<T> &v) {for (auto& a : v) { std::cout << a; if (&a != &v.back()) {std::cout << " ";} }std::cout << std::endl;}
template <class T> void print(std::vector<std::vector<T>> &vv) { for (auto& v : vv) { print(v); }}
void debug() { std::cerr << std::endl; }
template <class Head, class... Tail> void debug(Head&& head, Tail&&... tail) { std::cerr << head; if (sizeof...(tail) != 0) {std::cerr << " ";} print(std::forward<Tail>(tail)...); }
template <class T> void debug(std::vector<T> &v) {for (auto& a : v) { std::cerr << a; if (&a != &v.back()) {std::cerr << " ";} }std::cerr << std::endl;}
template <class T> void debug(std::vector<std::vector<T>> &vv) { for (auto& v : vv) { print(v); }}
inline bool inside(long long y, long long x, long long H, long long W) {return 0 <= y and y < H and 0 <= x and x < W; }
template<class T> inline double euclidean_distance(T y1, T x1, T y2, T x2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }
template<class T> inline double manhattan_distance(T y1, T x1, T y2, T x2) { return abs(x1 - x2) + abs(y1 - y2); }
template<typename T> T &chmin(T &a, const T &b) { return a = std::min(a, b); }
template<typename T> T &chmax(T &a, const T &b) { return a = std::max(a, b); }

template<class T> inline std::vector<T> unique(std::vector<T> v) {
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    return v;
}

// 初項s交差d長さnの数列の和
long long sum_of_arithmetic_progression(long long s, long long d, long long n) {
    return n * (2 * s + (n - 1) * d) / 2;
}

// xが2の階乗かどうか判定
bool is_power_of_two(long long x) {
    return !(x & (x - 1));
}

long long gcd(long long a, long long b) {
    if (b == 0) { return a; }
    return gcd(b, a % b);
}

long long gcd(std::vector<long long> &v) {
    long long ans = v[0];
    for (int i = 1; i < (int) v.size(); ++i) {
        ans = gcd(ans, v[i]);
    }
    return ans;
}

long long lcm(long long a, long long b) {
    long long g = gcd(a, b);
    return a / g * b;
}

const int INF = 1u << 30u;  // 1,073,741,824
const long long LINF = 1ull << 60u;
const double EPS = 1e-9;
const long double PI = acos(-1.0);
const std::vector<int> dy2 = {0, 1}, dx2 = {1, 0};
const std::vector<int> dy4 = {0, 1, 0, -1}, dx4 = {1, 0, -1, 0};
const std::vector<int> dy8 = {0, -1, 0, 1, 1, -1, -1, 1}, dx8 = {1, 0, -1, 0, 1, 1, -1, -1};

using namespace std;


bool ok(LL x, map<pair<int, int>, LL> &sum, vector<vector<int>> &graph) {
    int N = LEN(graph);
    auto number = make_v<LL>(N);
    fill_v(number, LINF);

    number[0] = x;

    queue<int> que;
    que.emplace(0);

    while (not que.empty()) {
        int u = que.front(); que.pop();
        FOE(v, graph[u]) {

            if (number[v] == LINF) {
                number[v] = sum[make_pair(u, v)] - number[u];
                if (number[v] <= 0) {
                    return false;
                }
                que.emplace(v);
            }
            else {
                if (number[u] + number[v] != sum[make_pair(u, v)]) {
                    return false;
                }
            }
        }
    }

    return true;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    LL N, M;
    cin >> N >> M;

    auto graph = make_v<int>(N, 0);
    map<pair<int, int>, LL> sum;
    FOR(i, 0, M) {
        LL U, V, S;
        cin >> U >> V >> S;
        U--; V--;
        graph[U].emplace_back(V);
        graph[V].emplace_back(U);
        sum[make_pair(U, V)] = S;
        sum[make_pair(V, U)] = S;
    }

    vector<pair<LL, LL>> number(N, make_pair(LINF, LINF));
    number[0] = make_pair(1, 0);

    LL fixed_x = LINF;

    queue<int> que;
    que.emplace(0);
    while (not que.empty()) {
        int u = que.front(); que.pop();

        FOE(v, graph[u]) {
            if (number[v].first == LINF) {
                number[v].first = 0 - number[u].first;
                number[v].second = sum[make_pair(u, v)] - number[u].second;
                que.emplace(v);
            }
            else {
                LL v_a, v_b;
                tie(v_a, v_b) = number[v];
                LL v_c, v_d;
                v_c = 0 - number[u].first;
                v_d = sum[make_pair(u, v)] - number[u].second;

                if (v_a + v_c == 0 and v_b + v_d == sum[make_pair(u, v)]) {
                    continue;
                }
                if (v_a == v_c and v_b == v_d) {
                    continue;
                }

                if ((v_a - v_c) == 0) {
                    print(0);
                    return 0;
                }

                LL x = (v_d - v_b) / (v_a - v_c);
                if (x <= 0 or (v_d - v_b) % (v_a - v_c) != 0) {
                    print(0);
                    return 0;
                }
                fixed_x = x;
            }
        }
    }

    // 1個
    if (fixed_x != LINF) {
        print(ok(fixed_x, sum, graph));
        return 0;
    }

    LL left = 0;
    LL right = LINF;
    FOR(u, 0, N) {
        LL a, b;
        tie(a, b) = number[u];

        if (a > 0) {
            chmax(left, (-b / a + 1));
        }
        else if (a < 0) {
            chmin(right, -b / a - 1);
        }
    }

    print(max(0LL, right - left + 1));

    return 0;
}