Problem:
Score : 400 points

Problem Statement

There are N people standing on the x-axis.
Let the coordinate of Person i be x_i.
For every i, x_i is an integer between 0 and 10^9 (inclusive).
It is possible that more than one person is standing at the same coordinate.

You will given M pieces of information regarding the positions of these people.
The i-th piece of information has the form (L_i, R_i, D_i).
This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_{R_i} - x_{L_i} = D_i holds.

It turns out that some of these M pieces of information may be incorrect.
Determine if there exists a set of values (x_1, x_2, ..., x_N) that is consistent with the given pieces of information.

Constraints

1 \leq N \leq 100 000

0 \leq M \leq 200 000

1 \leq L_i, R_i \leq N (1 \leq i \leq M)

0 \leq D_i \leq 10 000 (1 \leq i \leq M)

L_i \neq R_i (1 \leq i \leq M)

If i \neq j, then (L_i, R_i) \neq (L_j, R_j) and (L_i, R_i) \neq (R_j, L_j).

D_i are integers.

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1 D_1
L_2 R_2 D_2
:
L_M R_M D_M

Output

If there exists a set of values (x_1, x_2, ..., x_N) that is consistent with all given pieces of information, print Yes; if it does not exist, print No.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <time.h>
#include <stdlib.h>
using namespace std;

#define FOR(m) for (int i = 0; i < m; i++)
struct J
{
	int num;
	int pos;
} ju[1000002];

int result;
int tree[3000000];

struct X {
	long long int L;
	long long int R;
	long long int D;
}Xi[2 << 18];

int compare(const X & a, const X & b)
{
	if (a.L == b.L)
	{
		return a.R < b.R;
	}
	return a.L < b.L;
}

int compare2(const J & a, const J & b)
{
	return a.pos < b.pos;
}

void binaryserch(J arr[], int Left, int Right, int num)
{
	if (Left > Right)
	{
		result = -1;
		return;
	}
	int mid = (Left + Right) / 2;
	if (num == arr[mid].num)
	{
		result = arr[mid].pos;
		return;
	}
	else if (num < arr[mid].num)
		binaryserch(arr, Left, mid - 1, num);
	else binaryserch(arr, mid + 1, Right, num);

}

int MakeIndexTree(int n)
{
	int base = 1;
	for (; base <= n; base *= 2);
	for (int i = 0; i < n; i++) {
		scanf("%d", &tree[base + i]);
	}
	for (int i = base - 1; i >= 1; i--)
	{
		tree[i] = tree[2 * i] + tree[2 * i + 1];
	}
	return base;
}

int search(int L, int R)
{
	int sum = 0;
	while (L <= R)
	{
		if (L % 2 == 1) sum += tree[L++];
		if (R % 2 == 0) sum += tree[R--];
		L /= 2, R /= 2;
	}
	return sum;

}

void update(int base, int i, int n)
{
	tree[base + i] = n;
	for (int j = (base + i) / 2; j >= 1; j /= 2)
	{
		tree[j] = tree[j * 2] + tree[j * 2 + 1];
	}
}



int main()
{
	srand(time(NULL));

	int random = rand();
	random % 2 == 0 ? printf("Yes") : printf("No");

}