Interesting Array

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 l _i, r _i, q _i (1 ≤ l _i ≤ r _i ≤ n) meaning that value  should be equal to q _i.

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 l _i, r _i, q _i (1 ≤ l _i ≤ r _i ≤ n, 0 ≤ q _i < 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

Solution:

#include <cstring>
#include <vector>
#include  <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>
#include <cassert>

using namespace std;

const int N = 200010;

int from[N], to[N], num[N];
int s[N][30];
int a[N][30];
int sum[N][30];

int main() {
int n, m;
scanf("%d %d", &n, &m);
for (int i = 0; i <= n; i++) {
    for (int j = 0; j < 30; j++) {
      s[i][j] = 0;
    }
  }
  for (int k = 0; k < m; k++) {
    scanf("%d %d %d", from + k, to + k, num + k);
    from[k]--; to[k]--;
    for (int j = 0; j < 30; j++) {
      if (num[k] & (1 << j)) {
        s[from[k]][j]++;
        s[to[k] + 1][j]--;
      }
    }
  }
  for (int j = 0; j < 30; j++) {
    int bal = 0;
    sum[0][j] = 0;
    for (int i = 0; i < n; i++) {
      bal += s[i][j];
      a[i][j] = (bal > 0);
sum[i + 1][j] = sum[i][j] + a[i][j];
}
}
for (int k = 0; k < m; k++) {
    for (int j = 0; j < 30; j++) {
      if (!(num[k] & (1 << j))) {
        int get = sum[to[k] + 1][j] - sum[from[k]][j];
        int need = (to[k] + 1) - (from[k]);
        if (get == need) {
          puts("NO");
          return 0;
        }
      }
    }
  }
  puts("YES");
  for (int i = 0; i < n; i++) {
    int res = 0;
    for (int j = 0; j < 30; j++) {
      if (a[i][j]) {
        res += (1 << j);
      }
    }
    if (i > 0) printf(" ");
printf("%d", res);
}
printf("\n");
return 0;
}