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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

const int MaxN = 1e5 + 6.66;
vector<pair<int, int>> g[MaxN];
int mark[MaxN];
int mn[2];
vector<int> odd_c;
int st_c;

int odd_cycle(int v = 1, int p = 0, int s = 1) {
  mark[v] = s;
  s = s % 2 + 1;
  for(auto x : g[v]) {
    if(x.first == p) continue;
    if(mark[x.first] == 0) {
      int z = odd_cycle(x.first, v, s);
      if(z != -1) {
        if(z) {
          odd_c.push_back(x.second);
          if(z == v) return 0;
        }
        return z;
      }
    } else if(mark[x.first] != s) {
      odd_c.push_back(x.second);
      st_c = x.first;
      return x.first;
    }
  }
  return -1;
}

bool check_dfs(int v, int val) {
  if(val <= 0) return false;
  mark[v] = val;
  for(auto x : g[v]) {
    if(mark[x.first]) {
      if(x.second != mark[x.first] + val) return false;
    } else if(!check_dfs(x.first, x.second - val)) return false;
  }
  return true;
}

bool bi_dfs(int v, int val, bool sign) {
  mn[sign] = min(mn[sign], val);
  mark[v] = val;
  for(auto x : g[v]) {
    if(mark[x.first]) {
      if(x.second != mark[x.first] + val) return false;
      continue;
    }
    if(!bi_dfs(x.first, x.second - val, !sign)) return false;
  }
  return true;
}

int main() {
  int n, m; scanf("%d %d", &n, &m);
  for(int i = 0; i < m; i++) {
    int u, v, w; scanf("%d %d %d", &u, &v, &w);
    g[u].push_back({v, w});
    g[v].push_back({u, w});
  }
  if(~odd_cycle()) {
    long long sign = 1;
    long long sum = 0;
    for(auto x : odd_c) {
      sum += sign * x;
      sign *= -1;
    }
    memset(mark, 0, sizeof mark);
    printf("%d\n", sum % 2 == 0 && sum > 1 && check_dfs(st_c, sum / 2));
    return 0;
  }
  mn[0] = mn[1] = 2e9 + 7;
  memset(mark, 0, sizeof mark);
  if(!bi_dfs(1, 0, 0)) printf("0\n");
  else if(mn[1] <= 0 && mn[0] <= 0) printf("%d\n", 0);
  else printf("%d\n", max(0, mn[1] - mn[0] - 1));
  return 0;
}