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 l[100005];
int r[100005];
int q[100005];
int dp[100005];
int zeros[100005];
int ans[100005];
int main() {
  int n, m, pro = 1;
  cin >> n >> m;
  for (int i = 0; i < m; i++) cin >> l[i] >> r[i] >> q[i];
  bool chk = true;
  for (int i = 0; i < 30; i++) {
    memset(dp, 0, sizeof(dp));
    for (int j = 0; j < m; j++) {
      if (q[j] & pro) {
        dp[l[j]]++;
        dp[r[j] + 1]--;
      }
    }
    dp[0] = zeros[0] = 0;
    for (int j = 1; j <= n; j++) dp[j] += dp[j - 1];
    for (int j = 1; j <= n; j++) zeros[j] = zeros[j - 1] + (dp[j] == 0);
    for (int j = 0; j < m; j++) {
      if ((q[j] & pro) == 0) {
        if (zeros[r[j]] - zeros[l[j] - 1] == 0) chk = false;
      }
    }
    for (int i = 1; i <= n; i++) {
      if (dp[i]) ans[i] += pro;
    }
    pro *= 2;
  }
  if (!chk)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}
