482_B. 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 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
```python
import sys

input = sys.stdin.readline

n, m = map(int, input().split())
a = []
for _ in range(m):
    l, r, q = map(int, input().split())
    l -= 1
    r -= 1
    a.append((l, r, q))
res = [0] * n
bad = False
for i in range(30):
    events = [0] * (n + 1)
    for l, r, q in a:
        if q & (1 << i):
            events[l] += 1
            events[r + 1] -= 1
    c = 0
    for j in range(n):
        c += events[j]
        if c > 0:
            res[j] |= (1 << i)
    s = [0] * (n + 1)
    for j in range(n):
        s[j + 1] = s[j] + ((res[j] >> i) & 1)
    for l, r, q in a:
        if q & (1 << i) == 0:
            if s[r + 1] - s[l] == r - l + 1:
                bad = True
                break
if bad:
    print("NO")
else:
    print("YES")
    print(*res)
```