Problem: Consider placing N flags on a line. Flags are numbered through 1 to N.

Flag i can be placed on the coordinate X_i or Y_i. For any two different flags, the distance between them should be at least D.

Decide whether it is possible to place all N flags. If it is possible, print such a configulation.

Constraints

* 1 \leq N \leq 1000
* 0 \leq D \leq 10^9
* 0 \leq X_i < Y_i \leq 10^9
* All values in Input are integer.

Input

Input is given from Standard Input in the following format:


N D
X_1 Y_1
X_2 Y_2
\vdots
X_N Y_N


Output

Print `No` if it is impossible to place N flags.

If it is possible, print `Yes` first. After that, print N lines. i-th line of them should contain the coodinate of flag i.

Examples

Input

3 2
1 4
2 5
0 6


Output

Yes
4
2
0


Input

3 3
1 4
2 5
0 6


Output

No

Solution (in PYTHON3):

Solution:
N,D = map(int,input().split())
XY = [tuple(map(int,input().split())) for i in range(N)]

import sys
sys.setrecursionlimit(10**8)
class Scc:
    def __init__(self,n):
        self.n = n
        self.edges = []

    def add_edge(self,fr,to):
        assert 0 <= fr < self.n
        assert 0 <= to < self.n
        self.edges.append((fr, to))

    def scc(self):
        csr_start = [0] * (self.n + 1)
        csr_elist = [0] * len(self.edges)
        for fr,to in self.edges:
            csr_start[fr + 1] += 1
        for i in range(1,self.n+1):
            csr_start[i] += csr_start[i-1]
        counter = csr_start[:]
        for fr,to in self.edges:
            csr_elist[counter[fr]] = to
            counter[fr] += 1

        self.now_ord = self.group_num = 0
        self.visited = []
        self.low = [0] * self.n
        self.ord = [-1] * self.n
        self.ids = [0] * self.n
        def _dfs(v):
            self.low[v] = self.ord[v] = self.now_ord
            self.now_ord += 1
            self.visited.append(v)
            for i in range(csr_start[v], csr_start[v+1]):
                to = csr_elist[i]
                if self.ord[to] == -1:
                    _dfs(to)
                    self.low[v] = min(self.low[v], self.low[to])
                else:
                    self.low[v] = min(self.low[v], self.ord[to])
            if self.low[v] == self.ord[v]:
                while 1:
                    u = self.visited.pop()
                    self.ord[u] = self.n
                    self.ids[u] = self.group_num
                    if u==v: break
                self.group_num += 1
        for i in range(self.n):
            if self.ord[i] == -1: _dfs(i)
        for i in range(self.n):
            self.ids[i] = self.group_num - 1 - self.ids[i]

        groups = [[] for _ in range(self.group_num)]
        for i in range(self.n):
            groups[self.ids[i]].append(i)
        return groups

class TwoSat:
    def __init__(self,n=0):
        self.n = n
        self.answer = []
        self.scc = Scc(2*n)

    def add_clause(self, i:int, f:bool, j:int, g:bool):
        assert 0 <= i < self.n
        assert 0 <= j < self.n
        self.scc.add_edge(2*i + (not f), 2*j + g)
        self.scc.add_edge(2*j + (not g), 2*i + f)

    def satisfiable(self):
        g = self.scc.scc()
        for i in range(self.n):
            if self.scc.ids[2*i] == self.scc.ids[2*i+1]: return False
            self.answer.append(self.scc.ids[2*i] < self.scc.ids[2*i+1])
        return True

    def get_answer(self):
        return self.answer

ts = TwoSat(N)
for i in range(N-1):
    xi,yi = XY[i]
    for j in range(i+1,N):
        xj,yj = XY[j]
        if abs(xi - xj) < D:
            ts.add_clause(i, False, j, False)
        if abs(xi - yj) < D:
            ts.add_clause(i, False, j, True)
        if abs(yi - xj) < D:
            ts.add_clause(i, True, j, False)
        if abs(yi - yj) < D:
            ts.add_clause(i, True, j, True)
if not ts.satisfiable():
    print('No')
    exit()
print('Yes')
ans = []
for b,(x,y) in zip(ts.get_answer(), XY):
    if b:
        ans.append(x)
    else:
        ans.append(y)
print(*ans, sep='\n')