//struct of bign(big number)
struct bign {
    int d[1000];
    int len;
	bign()//構(gòu)造函數(shù)
	{
		this->len = 0;
		memset(d, 0, sizeof(d));
	}
};

bign change(const string str)//將整數(shù)轉(zhuǎn)換為begin c語(yǔ)言用const char* 
{
	bign a;
	a.len = str.size();//bign的長(zhǎng)度就是字符串的長(zhǎng)度 c語(yǔ)言用strlen()函數(shù)
	for (int i = 0; i < a.len; i++)
	{
		a.d[i] = str[a.len - i - 1] - '0';
	}
	return a;
}

bign add(bign a, bign b)
{
	bign c;
	int carry = 0;	//carry是進(jìn)位標(biāo)志
	for (int i = 0; i < a.len || i < b.len; i++) //以較長(zhǎng)的為界限 
	{
		int temp = a.d[i] + b.d[i] + carry;		//兩個(gè)對(duì)應(yīng)位與進(jìn)位相加 
		c.d[c.len++] = temp % 10;		//取個(gè)位數(shù)為該位的結(jié)果 
		carry = temp / 10;	//取十位數(shù)為新的進(jìn)位 
	}
	if (carry) //如果最后的進(jìn)位不為0，則直接賦給結(jié)果的最高位  
	{
		c.d[c.len++] = carry;
	}
	return c;
}

bign sub(bign a, bign b) //a - b
{
	bign c;
	for (int i = 0; i < a.len || i < b.len; i++) //以較長(zhǎng)的為界限
	{
		if (a.d[i] < b.d[i]) //不夠減 
		{
			a.d[i + 1]--;	//向高位借位 
			a.d[i] += 10; 	//向前位借10  
		}
		c.d[c.len++] = a.d[i] - b.d[i];		//減法結(jié)果為當(dāng)前位 
	}
	while (c.len >= 2 && c.d[c.len - 1] == 0)    //剩余的位數(shù)不小于十位，并且最高位是0
	{
		c.len--;	//去除高位的0，同時(shí)至少保留一位最低位 
	}
	return c;
}

bign multi(bign a, int b)
{
	bign c;
	int carry = 0;	//進(jìn)位
	for (int i = 0; i < a.len; i++)
	{
		int temp = a.d[i] * b + carry;
		c.d[c.len++] = temp % 10;		//個(gè)位作為該結(jié)果
		carry = temp / 10;	//高位部分作為新的進(jìn)位 
	}
 
	while (carry != 0) //和加法不一樣，乘法的進(jìn)位可能不止一位，因此用while 
	{
		c.d[c.len++] = carry % 10;
		carry /= 10;
	}
	return c;
}

bign divide(bign a, int b, int& r) //r為余數(shù) 
{
	bign c;
	c.len = a.len;//被除數(shù)的每一位和商的每一位是一一對(duì)應(yīng)的，因此先令長(zhǎng)度相等 
	for (int i = a.len - 1; i >= 0; i--)  //從高位開始 
	{
		r = r * 10 + a.d[i];		//和上一位遺留的余數(shù)組合
		if (r < b) c.d[i] = 0;		//不夠除，該位為0 
		else //夠除 
		{
			c.d[i] = r / b;	//商
			r = r % b;		//獲得新的余數(shù) 
		}
	}
	while (c.len >= 2 && c.d[c.len - 1] == 0)
	{
		c.len--; //去除高位的0，同時(shí)至少保留一位最低位 
	}
	return c;
}

void print(bign ans)
{
	int flag = 0;
	for (int i = ans.len - 1; i >= 0; i--)
	{
		if (ans.d[i] != 0)    //標(biāo)志位如果首位是0 則不輸出
		{
			flag = 1;
		}
		if (flag)
		{
			cout << ans.d[i];
		}
	}
	if (!flag)
		cout << 0;    //包含輸出0的情況
}

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn = 10001;
pair<int,int> nums[maxn];	//first 右手,second左手

struct bign{	//存123時(shí): d={3,2,1}llen =3; 
	int d[maxn];
	int len;
	bign(){
		memset(d,0,sizeof(d));
		len = 0;}
}res,sum,ans;	//res最終結(jié)果,sum累乘結(jié)果,ans比較變量

bign multi(bign a,int b){	//高精度大整數(shù)與低精度的乘法
	bign c;
	int carry = 0; //進(jìn)位
	for(int i=0;i<a.len;i++){
		int temp = a.d[i]*b + carry;
		//printf("%d*%d+%d\n",a.d[i],b,carry);
		c.d[c.len++] = temp%10;
		carry = temp / 10;
	} 
	while(carry!=0){
		c.d[c.len++] = carry%10;
		carry/=10;
	}
	return c;
}

bign divide(bign a,int b,int& r){	//高精度除以低精度,r為余數(shù) 
	bign c;
	c.len = a.len;
	r=0;
	for(int i=a.len-1;i>=0;i--){
		r = r*10 + a.d[i];
		if(r<b) c.d[i] = 0;
		else{
			c.d[i] = r/b;
			r %= b;
		}
	}
	while(c.len-1>=1&&c.d[c.len-1]==0){
		c.len--;
	}
	return c;
}

int compare(bign a,bign b){	//比較a和b的大小,a>b返回1,相等返回0,a<b返回-1 
	if(a.len>b.len) return 1;
	if(a.len<b.len) return -1;
	for(int i=a.len-1;i>=0;i--){
		if(a.d[i]>b.d[i]) return 1;
		if(a.d[i]<b.d[i]) return -1;
	}
	return 0;
}

void print(bign a){	//輸出大整數(shù) 
	for(int i=a.len-1;i>=0;i--)
		printf("%d",a.d[i]);
	return;
}

int cmp(pair<int,int> a,pair<int,int> b){
	return a.first*a.second<b.first*b.second;
}

int main(){
	int n;
	scanf("%d",&n);
	scanf("%d %d",&nums[0].first,&nums[0].second);
	for(int i=1;i<=n;i++)
		scanf("%d %d",&nums[i].first,&nums[i].second);
	sort(nums+1,nums+1+n,cmp);
	sum.d[0] = 1; sum.len = 1;
	for(int i=1;i<=n;i++){
		int r = 0;	//存余數(shù) 
		sum = multi(sum,nums[i-1].first);
		ans = divide(sum,nums[i].second,r);
		if(compare(ans,res)==1)
			res = ans;
	}
	print(res);
	return 0;
}