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 = 262144;
int n, m;
vector<int> p(N, 0);
void add(int pos, int a, int b, int l, int r, int v) {
  if (a >= r || b <= l) {
    return;
  } else if (a >= l && b <= r) {
    p[pos] |= v;
  } else {
    add(pos * 2, a, (a + b) / 2, l, r, v);
    add(pos * 2 + 1, (a + b) / 2, b, l, r, v);
  }
}
int query(int pos, int a, int b, int l, int r) {
  if (a >= r || b <= l) {
    return (1 << 30) - 1;
  } else if (a >= l && b <= r) {
    return p[pos];
  } else {
    return query(pos * 2, a, (a + b) / 2, l, r) &
           query(pos * 2 + 1, (a + b) / 2, b, l, r);
  }
}
int get(int x) {
  x += N / 2;
  int r = p[x];
  while (x >= 1) {
    r |= p[x];
    x /= 2;
  }
  return r;
}
int main() {
  cin >> n >> m;
  int l[100000], r[100000], q[100000];
  for (int i = 0; i < m; i++) {
    cin >> l[i] >> r[i] >> q[i];
    l[i]--;
    add(1, 0, N / 2, l[i], r[i], q[i]);
  }
  for (int i = 0; i < m; i++) {
    if (query(1, 0, N / 2, l[i], r[i]) != q[i]) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    cout << get(i) << " ";
  }
  cout << endl;
  return 0;
}
