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;
const int NMAX = 1e5;
const int LOG = 30;
int v[LOG + 1][NMAX + 2];
int xx[NMAX + 2], yy[NMAX + 2], qq[NMAX + 2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int N, Q;
  cin >> N >> Q;
  for (int i = 1; i <= Q; ++i) {
    int x, y, val;
    cin >> x >> y >> val;
    xx[i] = x;
    yy[i] = y;
    qq[i] = val;
    for (int j = 0; j <= LOG; ++j) {
      if (val & (1 << j)) {
        v[j][x]++;
        v[j][y + 1]--;
      }
    }
  }
  for (int j = 0; j <= LOG; ++j)
    for (int i = 1; i <= NMAX; ++i) v[j][i] += v[j][i - 1];
  for (int j = 0; j <= LOG; ++j) {
    for (int i = 1; i <= NMAX; ++i) {
      if (v[j][i])
        v[j][i] = 1;
      else
        v[j][i] = 0;
      v[j][i] += v[j][i - 1];
    }
  }
  for (int i = 1; i <= Q; ++i) {
    int x = xx[i];
    int y = yy[i];
    int val = qq[i];
    int actual = 0;
    for (int j = 0; j <= LOG; ++j) {
      if (v[j][y] - v[j][x - 1] == (y - x + 1)) actual += (1 << j);
    }
    if (actual != val) {
      cout << "NO\n";
      return 0;
    }
  }
  cout << "YES\n";
  for (int i = 1; i <= N; ++i) {
    int val = 0;
    for (int j = 0; j <= LOG; ++j) {
      if (v[j][i] - v[j][i - 1] > 0) val += (1 << j);
    }
    cout << val << " \n"[i == N];
  }
  return 0;
}
