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, k;
int curr[30];
int ar[100003];
int tot[100003][30];
int add[100003][30];
int query[100003][3];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < k; i++) {
    scanf("%d%d%d", &query[i][0], &query[i][1], &query[i][2]);
    for (int j = 0; j < 30; j++)
      if (query[i][2] & (1 << j)) {
        add[query[i][0]][j]++;
        add[query[i][1] + 1][j]--;
      }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 30; j++) {
      curr[j] += add[i][j];
      tot[i][j] = tot[i - 1][j];
      if (curr[j]) {
        ar[i] |= (1 << j);
        tot[i][j]++;
      }
    }
  }
  for (int i = 0; i < k; i++) {
    int x = query[i][0], y = query[i][1], z = query[i][2], val = 0;
    for (int j = 0; j < 30; j++)
      if (tot[y][j] - tot[x - 1][j] == y - x + 1) val |= (1 << j);
    if (val != z) {
      printf("NO\n");
      exit(0);
    }
  }
  printf("YES\n");
  for (int i = 1; i <= n; i++) printf("%d ", ar[i]);
  return 0;
}
