Problem: We often go to supermarkets to buy some fruits or vegetables, and on the tag there prints the price for a kilo. But in some supermarkets, when asked how much the items are, the clerk will say that a yuan for b kilos (You don't need to care about what "yuan" is), the same as a/b yuan for a kilo.

Now imagine you'd like to buy m kilos of apples. You've asked n supermarkets and got the prices. Find the minimum cost for those apples.

You can assume that there are enough apples in all supermarkets.

Input

The first line contains two positive integers n and m (1 ≤ n ≤ 5 000, 1 ≤ m ≤ 100), denoting that there are n supermarkets and you want to buy m kilos of apples.

The following n lines describe the information of the supermarkets. Each line contains two positive integers a, b (1 ≤ a, b ≤ 100), denoting that in this supermarket, you are supposed to pay a yuan for b kilos of apples.

Output

The only line, denoting the minimum cost for m kilos of apples. Please make sure that the absolute or relative error between your answer and the correct answer won't exceed 10^{-6}.

Formally, let your answer be x, and the jury's answer be y. Your answer is considered correct if \frac{|x - y|}{max{(1, |y|)}} ≤ 10^{-6}.

Examples

Input

3 5
1 2
3 4
1 3


Output

1.66666667


Input

2 1
99 100
98 99


Output

0.98989899

Note

In the first sample, you are supposed to buy 5 kilos of apples in supermarket 3. The cost is 5/3 yuan.

In the second sample, you are supposed to buy 1 kilo of apples in supermarket 2. The cost is 98/99 yuan.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long double ldb;
const int maxn = 1 << 21;
const int N = int(1e5);
const long long mod = int(1e9) + 7;
const long long inf = 1e18;
const double PI = 3.1415926536;
inline int Set(int n, int pos) { return n = n | 1 << pos; }
inline bool check(int n, int pos) { return n & 1 << pos; }
inline int Reset(int n, int pos) { return n = n & ~(1 << pos); }
inline int lastbit(int n) { return n & (-n); }
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return (a * b) / gcd(a, b); }
inline long long power(long long a, long long b) {
  long long ans = 1;
  a = a % mod;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return ans;
}
inline long long invmod(long long a) {
  long long ans = power(a, mod - 2);
  return ans;
}
inline long long Egcd(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = Egcd(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
inline int bs(int arr[], int l, int r, int item) {
  if (l == r) {
    if (arr[l] == item)
      return r;
    else
      return -1;
  }
  if (item <= arr[(l + r) / 2])
    return bs(arr, l, (l + r) / 2, item);
  else
    return bs(arr, (l + r) / 2 + 1, r, item);
}
inline void fastin(int &x) {
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return;
}
int leap[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int nleap[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
vector<pair<int, int> > v;
bool cmp(pair<int, int> a, pair<int, int> b) {
  double r1 = (double)a.first / a.second;
  double r2 = (double)b.first / b.second;
  return r1 < r2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    v.push_back(make_pair(a, b));
  }
  sort(v.begin(), v.end(), cmp);
  double ans = (double)v[0].first / v[0].second;
  ans *= m;
  cout << fixed << setprecision(12) << ans;
}