Codeforces 482B Interesting Array

Description

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 should be equal T O 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 are represented as "&", in Pascal-as "and".

Input

The first line contains-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 <) describing the i-th limit.

Output

If the interesting array exists, in the first line print "YES" (without the quotes) and the second line print n integer S a[1], a[2], ..., A[n] (0≤a[i] <) decribing the interesting array. If There is multiple answers, print any of them.

If the interesting array doesn ' t exist, print "NO" (without the quotes) in the.

Sample Input Input

3 1
1) 3 3

Output

YES
3 3 3

Input

3 2
1 3 3
1 3 2

Output

NO

/** Test Instructions: give n number and m operation, each operation shape such as: l R x, means A[l] & A[l+1] & ... &a[r] = x;
                    Q: Is there a sequence that satisfies the given operation?
                There are: output Yes, and output this sequence.
    No: Output No.
                Analysis: Consider A[i], for all operations that contain it, if the operation is X, the binary of X is 1 bits, and A[i] is also 1.
                The other bits are 0 sure to be satisfied.
                So what we can do with all x or, what we get is a[i].
                The segment tree interval operation is not much to say.
   Finally, the test sequence meets the requirements. */#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&LT;&LT;1);
if (r>m) update (L,R,X,M+1,R,RT&LT;&LT;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;
 }