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 N = 100010;
int n, m;
int a[N];
int l[N], r[N], q[N];
int sum[N];
int T[4 * N];
void build(int p, int l, int r) {
  if (l == r) {
    T[p] = a[l];
    return;
  }
  int m = (l + r) >> 1;
  build(p << 1, l, m);
  build(p << 1 | 1, m + 1, r);
  T[p] = T[p << 1] & T[p << 1 | 1];
}
int get(int p, int l, int r, int u, int v) {
  if (v < l || r < u) return INT_MAX;
  if (u <= l && r <= v) return T[p];
  int m = (l + r) >> 1;
  int t1 = get(p << 1, l, m, u, v);
  int t2 = get(p << 1 | 1, m + 1, r, u, v);
  return (t1 & t2);
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; ++i) {
    scanf("%d %d %d", &l[i], &r[i], &q[i]);
  }
  for (int bit = 0; bit <= 30; ++bit) {
    for (int i = 0; i <= n; ++i) sum[i] = 0;
    for (int i = 0; i < m; ++i) {
      if (q[i] & (1 << bit)) {
        sum[l[i]]++;
        sum[r[i] + 1]--;
      }
    }
    for (int i = 1; i <= n; ++i) sum[i] += sum[i - 1];
    for (int i = 1; i <= n; ++i)
      if (sum[i]) a[i] |= (1 << bit);
  }
  build(1, 1, n);
  for (int i = 0; i < m; ++i) {
    int value = get(1, 1, n, l[i], r[i]);
    if (value != q[i]) {
      puts("NO");
      return 0;
    }
  }
  puts("YES");
  for (int i = 1; i <= n; ++i) printf("%d ", a[i]);
  return 0;
}
