We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi. 

Your task is to find any interesting array of n elements or state that such array doesn't exist.

Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".

Input

The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.

Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.

Output

If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.

If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.

Examples

Input

3 1
1 3 3


Output

YES
3 3 3


Input

3 2
1 3 3
1 3 2


Output

NO
#include <bits/stdc++.h>
using namespace std;
int n, m, t1, t2, t3, arr[100005];
vector<pair<pair<int, int>, int> > all;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie();
  cin >> n >> m;
  for (int i = 0; i < m && cin >> t1 >> t2 >> t3; i++)
    all.push_back({{t1, t2}, t3});
  for (int b = 0; b < 31; b++) {
    vector<int> par(n + 5, 0);
    for (auto u : all)
      if ((u.second >> b) & 1) par[u.first.first]++, par[u.first.second + 1]--;
    for (int i = 1; i <= n; i++) par[i] += par[i - 1];
    for (int i = 0; i <= n; i++) par[i] = min(par[i], 1);
    for (int i = 1; i <= n; i++) par[i] += par[i - 1];
    for (auto u : all)
      if (((u.second >> b) & 1) == 0 &&
          par[u.first.second] - par[u.first.first - 1] ==
              u.first.second - u.first.first + 1)
        return cout << "NO", 0;
    for (int i = 1; i <= n; i++)
      if (par[i] - par[i - 1]) arr[i] |= (1 << b);
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << arr[i] << " ";
  return 0;
}
