human
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.

The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island i has coordinates [li, ri], besides, ri < li + 1 for 1 ≤ i ≤ n - 1.

To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i + 1)-th islads, if there are such coordinates of x and y, that li ≤ x ≤ ri, li + 1 ≤ y ≤ ri + 1 and y - x = a. 

The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.

Input

The first line contains integers n (2 ≤ n ≤ 2·105) and m (1 ≤ m ≤ 2·105) — the number of islands and bridges.

Next n lines each contain two integers li and ri (1 ≤ li ≤ ri ≤ 1018) — the coordinates of the island endpoints.

The last line contains m integer numbers a1, a2, ..., am (1 ≤ ai ≤ 1018) — the lengths of the bridges that Andrewid got.

Output

If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b1, b2, ..., bn - 1, which mean that between islands i and i + 1 there must be used a bridge number bi. 

If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.

Examples

Input

4 4
1 4
7 8
9 10
12 14
4 5 3 8

Output

Yes
2 3 1 

Input

2 2
11 14
17 18
2 9

Output

No

Input

2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999

Output

Yes
1 

Note

In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge...
# Sta
Can you provide a Python code to determine whether the bridges he got are enough to connect each pair of adjacent islands?
assistant
#Here is Python code to determine whether the bridges he got are enough to connect each pair of adjacent islands.
ndard libraries
import unittest
import sys
import re

# Additional libraries
import heapq

###############################################################################
# Fug Class
###############################################################################

class Fug:
    """ Fug representation """

    def __init__(self, args):
        """ Default constructor """
        self.list = args[0]
        self.alist = args[1]
        self.gn = len(self.list) - 1

        # Sorted list of bridges
        self.asrt = sorted((n, i) for i, n in enumerate(self.alist))

        # List of gaps between islands
        self.gaps = []
        prevli = self.list[0]
        for i in range(self.gn):
            li = self.list[i+1]
            min = li[0] - prevli[1]
            max = li[1] - prevli[0]
            self.gaps.append((min, max, i))
            prevli = li

        # Sorted list of gaps between islands
        self.gsrt = sorted(self.gaps)

        self.gmin = [n[0] for n in self.gsrt]
        self.result = [None]*self.gn
        self.heap = []

    def iterate(self):

        j = 0
        for (b, i) in self.asrt:

            # Traverse gmin array
            while j < self.gn and self.gmin[j] <= b:
                it = self.gsrt[j]
                heapq.heappush(self.heap, (it[1], it[0], it[2]))
                j += 1

            # Update result and remove the element from lists
            if self.heap:
                (mmax, mmin, mi) = self.heap[0]
                if mmin <= b and mmax >= b:
                    self.result[mi] = i + 1
                    heapq.heappop(self.heap)

            yield

    def calculate(self):
        """ Main calcualtion function of the class """

        for it in self.iterate():
            pass

        for n in self.result:
            if n is None:
                return "No"
        answer = " ".join([str(n) for n in self.result])

        return "Yes\n" + answer

###############################################################################
# Executable...