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
# by the authority of GOD     author: manhar singh sachdev #

import os,sys
from io import BytesIO, IOBase

def main():
    n,m = map(int,input().split())
    dp = [[0]*30 for _ in range(n+2)]
    op = []
    for _ in range(m):
        op.append(tuple(map(int,input().split())))
        l,r,q = op[-1]
        mask,cou = 1,29
        while mask <= q:
            if mask&q:
                dp[l][cou] += 1
                dp[r+1][cou] -= 1
            cou -= 1
            mask <<= 1
    ans = [[0]*30 for _ in range(n)]
    for i in range(30):
        a = 0
        for j in range(n):
            a += dp[j+1][i]
            dp[j+1][i] = dp[j][i]
            if a:
                ans[j][i] = 1
                dp[j+1][i] += 1
    for i in op:
        l,r,q = i
        mask = 1
        for cou in range(29,-1,-1):
            if not mask&q and dp[r][cou]-dp[l-1][cou] == r-l+1:
                print('NO')
                return
            mask <<= 1
    for i in range(n):
        ans[i] = int(''.join(map(str,ans[i])),2)
    print('YES')
    print(*ans)


# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
    main()