Problem:
Score : 300 points

Problem Statement

Hearing that energy drinks increase rating in those sites, Takahashi decides to buy up M cans of energy drinks.

There are N stores that sell energy drinks. In the i-th store, he can buy at most B_i cans of energy drinks for A_i yen (the currency of Japan) each.

What is the minimum amount of money with which he can buy M cans of energy drinks?

It is guaranteed that, in the given inputs, a sufficient amount of money can always buy M cans of energy drinks.

Constraints

All values in input are integers.

1 \leq N, M \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^5

B_1 + ... + B_N \geq M

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the minimum amount of money with which Takahashi can buy M cans of energy drinks.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.

Solution in C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#pragma warning(disable:4996)
#pragma GCC diagnostic ignored "-Wunused-result"
#pragma GCC diagnostic ignored "-Wwrite-strings"
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#define MAX_STRING 100
#define MAX_STORES 100000

int main() {

	int  nPriceList[MAX_STORES];
	int  nStockList[MAX_STORES];
	// i番目の店の売価はnPriceList[i]で、在庫はnStockList[i]である。
	// ただし、売価が同じ店は1レコードに集約する。
	
	char sInput[MAX_STRING]="";
	char* psInput=NULL;
	int i=0;
	int j=0;
	
	int nStores=0;
	int nTargetBottles=0;
	int nTempPrice=0;
	int nMaxPrice=0;
	int nTempStock=0;
	long long nTotalPrice=0;
	long long nTempTotal=0;
	int nPresentStoresCount=0;
	int nStoresIndex=0;
	int nSamePrice=0;
	int nFound=0;
	
	for(i=0;i<MAX_STORES;i++){
		nPriceList[i]=0;
		nStockList[i]=0;
	}
//	printf("debug 0\n");
	
	// read line 1
	gets(sInput);
	psInput = strtok(sInput," ");
	nStores=atoi(psInput);
	psInput = strtok(NULL," ");
	nTargetBottles=atoi(psInput);

//	printf("debug 1\n");
	
	//read line 2 (and lator)
	for(i=0;i<nStores;i++){
		gets(sInput);
		psInput = strtok(sInput," ");
		nTempPrice=atoi(psInput);
		psInput = strtok(NULL," ");
		nTempStock=atoi(psInput);

		if(nMaxPrice<nTempPrice){
			nMaxPrice=nTempPrice;
		}
		
		// 過去に同じ価格の店があったか調べる
		if(nPresentStoresCount>0){
			nSamePrice=0;
			for(j=0;j<nPresentStoresCount;j++){
				if(nPriceList[j]==nTempPrice){
					nSamePrice=1;
					nStoresIndex=j;
					break;
				}
			}			
		}

		// あれば過去の店とレコードを統合する　なければ新規追加
		if(nSamePrice){
			nStockList[nStoresIndex]=nStockList[nStoresIndex]+nTempStock;
		} else {
			nStoresIndex=nPresentStoresCount;
			nPriceList[nStoresIndex]=nTempPrice;
			nStockList[nStoresIndex]=nTempStock;
			nPresentStoresCount++;
		}
	}
	
	// calc total price
	for(nTempPrice=1;nTempPrice<=nMaxPrice && nTargetBottles>0;nTempPrice++){
		nFound=0;
		for(j=0;j<nPresentStoresCount;j++){
			if(nPriceList[j]==nTempPrice){
				nFound=1;
				nStoresIndex=j;
				break;
			}
		}
		
		if(nFound){
			if(nTargetBottles>=nStockList[nStoresIndex]){
				nTempTotal=(long long)nPriceList[nStoresIndex]*nStockList[nStoresIndex];
				nTotalPrice=nTotalPrice+nTempTotal;
				nTargetBottles=nTargetBottles-nStockList[nStoresIndex];
			} else {
				// buy only bottles which we need
				nTempTotal=(long long)nPriceList[nStoresIndex]*nTargetBottles;
				nTotalPrice=nTotalPrice+nTempTotal;
				nTargetBottles=0;
				
			}
		}
	}
	
	printf("%lld\n",nTotalPrice);

    return 0;
}