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;
ifstream in("file.in");
ofstream out("file.out");
struct comp {
  bool operator()(int a, int b) { return a > b; }
};
int n, m, p[35][100005], q[35][100005], l[100005], r[100005], last[35][100005],
    first, k;
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> l[i] >> r[i] >> first;
    for (int j = 0; j <= 30; j++) {
      q[j][i] = first % 2;
      first /= 2;
      if (q[j][i] == 1) {
        p[j][l[i]]++;
        p[j][r[i] + 1]--;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 30; j++) {
      p[j][i] += p[j][i - 1];
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 30; j++) {
      if (p[j][i])
        p[j][i] = 1, last[j][i] = last[j][i - 1];
      else
        p[j][i] = 0, last[j][i] = i;
    }
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j < 30; j++) {
      if (!q[j][i] && last[j][r[i]] < l[i]) {
        cout << "NO";
        return 0;
      }
    }
  }
  cout << "YES\n";
  for (int i = 1; i <= n; i++) {
    first = 0, k = 1;
    for (int j = 0; j <= 30; j++) {
      first += p[j][i] * k;
      k *= 2;
    }
    cout << first << ' ';
  }
}
