CodeForces 482B Interesting Array

Description

We'll call an array of n non-negative integersa[1], a[2], ..., a[n]interesting, if it meetsm constraints. The i-th of the m constraints consists of three integersl_i,r_i,q_i (1 ≤ l_i ≤ r_i ≤ n) meaning that value should be equal toq_i.

Your task is to find any interesting array ofn elements or state that such array doesn't exist.

Expression x&y means the bitwise AND of numbersx 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 integersl_i,r_i,q_i (1 ≤ l_i ≤ r_i ≤ n,0 ≤ q_i < 2³⁰) describing thei-th limit.

Output

If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line printn integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 2³⁰) decribing theinteresting 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.

Sample Input

Input

3 11 3 3

Output

YES3 3 3

Input

3 21 3 31 3 2

Output

NO

/**    题意：给n个数和m个操作，每个操作形如： l r x, 表示a[l] & a[l+1] & ... &a[r] = x;                问：是否存在这样一个序列满足所给操作？                    有：输出YES，并输出这个序列。                没有：输出NO。    分析：考虑a[i],对于所有包含它的操作，如果取与操作后为x，                则x的二进制为1的位上，a[i]也为1。                其他位都为0肯定是满足的。                那么我们可以对所有x取或，得到的就是a[i]。                线段树区间操作就不多说了。                最后，检验序列是否符合要求。   */#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<set>using namespace std;const int N = 100001;int ans[N],mak[N<<2];int ql[N],qr[N],qx[N];void pushdown(int rt){    if(!mak[rt]) return ;    mak[rt<<1]|=mak[rt];    mak[rt<<1|1]|=mak[rt];    mak[rt] = 0;}void update(int L, int R, int x, int l, int r, int rt){    if(L<=l && R>=r)    {        mak[rt]|=x;        return;    }    int m = l+r>>1;    if(L<=m) update(L,R,x,l,m,rt<<1);    if(R>m)   update(L,R,x,m+1,r,rt<<1|1);}void solve(int l, int r, int rt){    if(l==r)    {        ans[l] = mak[rt];        return ;    }    int m = l+r>>1;    pushdown(rt);    solve(l,m,rt<<1);    solve(m+1,r,rt<<1|1);}void build(int l, int r, int rt){    if(l==r)    {        mak[rt] = ans[l];        return ;    }    int m = l+r>>1;    build(l,m,rt<<1);    build(m+1,r,rt<<1|1);    mak[rt] = mak[rt<<1]&mak[rt<<1|1];}int query(int L, int R, int l, int r, int rt){    if(L<=l && R>=r) return mak[rt];    int m = l+r>>1;    int s = (1<<30)-1;    if(L<=m) s&=query(L,R,l,m,rt<<1);    if(R>m)   s&=query(L,R,m+1,r,rt<<1|1);    return s;}int main(){    int i,j,k,m,n;    while(scanf("%d %d",&n,&m)==2)    {        memset(mak,0,sizeof(mak));        for(i=1; i<=m; i++)        {            scanf("%d%d%d",ql+i,qr+i,qx+i);            update(ql[i],qr[i],qx[i],1,n,1);        }        solve(1,n,1);        build(1,n,1);        bool ok = true;        for(i=1; i<=m; i++)        {            int a = query(ql[i],qr[i],1,n,1);            if(a!=qx[i])            {                ok = false;                break;            }        }        if(!ok)        {            puts("NO");            continue;        };        puts("YES");        for(i=1; i<n; i++) printf("%d ",ans[i]);        printf("%d\n",ans[n]);    }    return 0;}