# 在 m 批客人中选择 n 批上 n 个桌子
n, m = [int(x) for x in input().split()]
# 每个桌子可容纳的最大人数
table = [int(x) for x in input().split()]
# 分别表示第 i 批客人的人数和预计消费金额
cus = [[int(x) for x in input().split()] for i in range(m)]# 将桌子按可容纳人数从小到大排列
table.sort()
# 按每批顾客的人数从小到大排序
cus = sorted(cus, key=lambda x: x[0])
# 总金额
money = 0for i in range(n):# 盛放第 i 个可接受的顾客批 jtemp = []# 注意 range 中要用 cus 的 length 时更新for j in range(len(cus)):if cus[j][0] > table[i]:breaktemp.append(cus[j])# 按消费金额排序temp = sorted(temp, key=lambda x: x[1])if temp:money += temp[-1][1]# 此批客人已就坐cus.remove()
print(money)

'''
题目：一只青蛙一次可以跳上1级台阶，也可以跳上2级……它也可以跳上n级。求该青蛙跳上一个n级的台阶总共有多少种跳法
''''''
因为n级台阶，第一步有n种跳法：跳1级、跳2级、到跳n级
跳1级，剩下n-1级，则剩下跳法是f(n-1)
跳2级，剩下n-2级，则剩下跳法是f(n-2)
所以f(n)=f(n-1)+f(n-2)+...+f(1)
因为f(n-1)=f(n-2)+f(n-3)+...+f(1)
所以f(n)=2*f(n-1)
然后求解这个无穷级数的和，正确答案应该是：每次至少跳一个，至多跳n个，一共有f(n)=2n-1种跳法
29ms
5632k
'''# -*- coding:utf-8 -*-
class Solution:def jumpFloorII(self, number):# write code herereturn 2**(number-1)

'''
1.长度超过8位
2.包括大小写字母.数字.其它符号,以上四种至少三种
3.不能有相同长度超2的子串重复
说明:长度超过2的子串
'''# import re, sys
#
# for i in sys.stdin.readlines():
#     print("OK" if len(i.strip()) > 8 and sum(
#         [1 if re.search(r"[A-Z]", i.strip()) else 0, 1 if re.search(r"[a-z]", i.strip()) else 0,
#          1 if re.search(r"[0-9]", i.strip()) else 0, 1 if re.search(r"[^0-9a-zA-Z]", i.strip()) else 0]) > 2 and sum(
#         map(lambda c: i.strip().count(i.strip()[c:c + 3]) > 1, range(1, len(i.strip()) - 3))) == 0 else "NG")# 略微思考会发现，只需要判断长度为3的子串是否出现即可。因为假设子串长度为4的出现，则一定包括了长度为3的子串。同时需要注意，
# 本题说的子串指包括了部分子串重叠的情况，
# 例如Qw11111*ed这个是不能通过的。再就是需要注意，判断子串的时候只需要判断到len(str)-3就行了。import systry:# 大小写，字母，def panchar(sss):standard = [0] * 4for i in sss:# print(i)# 0# 2# 1# A# b# print(len(sss))# 数字if i.isdigit():standard[0] = 1# print(i.isdigit())# 小写if i.islower():standard[1] = 1# 大写if i.isupper():standard[2] = 1# 全都是字母，数字，空格if not i.isalpha() and not i.isdigit() and not i.isspace():standard[3] = 1if sum(standard) >= 3:return Falsereturn True# 不能有相同长度超 2 的字串重复def zichuan(sss):for i in range(len(sss) - 3):zichuan_1 = sss[i: i + 3]zichuan_2 = sss[i + 1::]if zichuan_1 in zichuan_2:return Truereturn Falseresult = []while True:line = sys.stdin.readline().strip()if line == '':breakif len(line) <= 8:result.append('NG')# 大小写字母.数字.其它符号elif panchar(line):result.append('NG')elif zichuan(line):result.append('NG')else:result.append('OK')for i in result:print(i)
except:pass# # 循环输入，try catch
# while True:
#     try:
#         x = input().split()
#
#
#     except:
#         pass

# https://blog.csdn.net/u012193416/article/details/78790448
# # 冒泡排序
# # 时间复杂度  O(n**2) 空间复杂度  O(1)
# x = [int(i) for i in input().split(',')]
#
# # print(x)
#
# def mpsort(x):
#     n = len(x)
#     # print(n)
#     for i in range(n - 1):
#         for j in range(0, n - i - 1):
#             # print(x[j])
#             if x[j] > x[j + 1]:
#                 x[j], x[j + 1] = x[j + 1], x[j]
#     return x
#
# print(mpsort(x))# # 选择排序
# # 时间复杂度 O(n**2) 空间复杂度  O(1)
# x = [int(i) for i in input().split(',')]
#
# def xzsort(x):
#     n = len(x)
#     for i in range(n - 1):
#         min = i
#         for j in range(i + 1, n):
#             if x[j] < x[min]:
#                 min = j
#         x[i], x[min] = x[min], x[i]
#     return x
#
# print(xzsort(x))# # 插入排序
# # 时间复杂度 O(n**2) 空间复杂度  O(1)
# x = [int(i) for i in input().split(',')]
#
# def crsort(x):
#     n = len(x)
#     for i in range(1, n):
#         j = i
#         while j > 0:
#             if x[j] < x[j - 1]:
#                 x[j], x[j - 1] = x[j - 1], x[j]
#                 j -= 1
#             else:
#                 break
#     return x
#
# print(crsort(x))# # 希尔排序
# # 时间复杂度 O(nlogn)-O(n**2) 空间复杂度  O(1)
# x = [int(i) for i in input().split(',')]
#
# def shellsort(x):
#     n = len(x)
#     gap = n // 2
#
#     while gap > 0:
#         for i in range(gap, n):
#             j = i
#             while j > 0:
#                 if x[j] < x[j - gap]:
#                     x[j], x[j - gap] = x[j - gap], x[j]
#                     j -= gap
#                 else:
#                     break
#         gap //= 2
#     return x
#
# print(shellsort(x))# # 快速排序
# # 时间复杂度 O(nlogn) 空间复杂度  O(logn)-O(n)
# x = [int(i) for i in input().split(',')]
#
# def kpsort(x, first, last):
#     font = first
#     end = last
#     middle = x[first]
#
#     if first >= last:
#         return
#
#     while font < end:
#         while font < end and x[font] <= middle:
#             font += 1
#         x[end] = x[font]
#
#         while font < end and x[end] > middle:
#             end -= 1
#         x[font] = x[end]
#
#     x[font] = middle
#
#     kpsort(x, first, font - 1)
#     kpsort(x, font + 1, last)# 归并排序
# 时间复杂度 O(nlogn)  空间复杂度  O(N)
x = [int(i) for i in input().split(',')]def gbsort(x):length = len(x)if length <= 1:return xmid = length // 2left = gbsort(x[:mid])right = gbsort(x[mid:])left_point, right_pointer = 0, 0result = []while left_point < len(left) and right_pointer < len(right):if left[left_point] <= right[right_pointer]:result.append(left[left_point])left_point += 1else:result.append(right_pointer)right_pointer += 1result += left[left_point:]result += right[right_pointer]return resultprint(gbsort(x))

# 指针给的是节点值
class Node():def __init__(self, value=None):self.value = valueself.next = Nonedef insertnum(head, num):node = Node(num)if head == None:node.next = nodereturn nodenode = headpre = nodecur = node.nextwhile cur != head:if pre.value > num and cur.value <= num:breakpre = pre.nextcur = cur.next# num 小于节点值，pre只跑到最后一个节点，node跑道头结点pre.next = nodenode.next = cur# 是按顺序来的，返回的是 head 或者 node ，是有顺序决定的return head if head.value < num else nodenode = Node()
node.insertnum([[1, 2, 3], 5])