Problem:
Score : 100 points

Problem Statement

Ringo Kingdom Congress is voting on a bill.

N members are present, and the i-th member (1 ≤ i ≤ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box.

After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected.

In order for the bill to pass, at least how many members must be in favor of it?

Constraints

1 ≤ N ≤ 10^5

1 ≤ P ≤ 100

1 ≤ w_i ≤ 10^9

1 ≤ b_i ≤ 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N P
w_1 b_1
w_2 b_2
:
w_N b_N

Output

Print the minimum number of members in favor of the bill required for passage.

Sample Input 1

4 75
1 1
1 1
1 1
1 1

Sample Output 1

3

This is a "normal" vote, where each of the four members has one white ballot and one blue ballot. For the bill to pass, at least 75 percent of the four, that is, at least three must be in favor of it.

Sample Input 2

4 75
1 1
1 1
1 1
100 1

Sample Output 2

1

The "yes" vote of the member with 100 white ballots alone is enough to pass the bill.

Sample Input 3

5 60
6 3
5 9
3 4
7 8
4 7

Sample Output 3

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/*
オーダー
10**6	余裕を持って間に合う	
10**7	おそらく間に合う	余裕を持って間に合う
10**8	非常にシンプルな処理でない限り厳しい	おそらく間に合う
10**9		非常にシンプルな処理でない限り厳しい
logn	:OK
n		:10^7
nlogn	:10^6
n**2	:10^4
n**3	:300
2**n	:20
n!		:10



// 各桁の和を計算する関数
int findSumOfDigits(int n) {
  int sum = 0;
  while (n > 0) { // n が 0 になるまで
    sum += n % 10;
    n /= 10;
  }
  return sum;
}

sort(a, a + N, greater<int>());  // a[0:N] を大きい順にソート

int num[110] = {0};  // バケット
for (int i = 0; i < N; ++i) {
    num[d[i]]++;  // d[i] が 1 個増える
}

map<string, int> mp; // 連想配列 map<キー型, 値型> オブジェクト名
for (int i = 0; i < N; ++i) {
    auto itr = mp.find(s[i]);        // s[i] が設定されているか？
    if(itr != mp.end() ) {
         mp[s[i]] += 1;
    }
    else {
         mp[s[i]] += 1 ;
    }
}
stack<int> s;	//intをデータとするスタックを用意
s.push(1);		//{} -> {1}
printf("%d\n", s.top());	// 3
s.pop();

queue<int> que;	//intをデータとするキューを用意
que.push(1);		//{} -> {1}
printf("%d\n", que.front());	// 1
que.pop();
*/

#include <iostream>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <bitset>
#include <string.h>

using namespace std;

// #define for(i,a,b) for (int i=(a);i<(b);++i)
typedef long long ll;
typedef pair<ll,ll> P;
#define REP(i,n) for(long long i = 0; i < (long long)(n); i++)
#define pb push_back // vectorに要素追加
#define INF (ll)1e18
// int 
// // 各桁の和を計算する関数
// int findSumOfDigits(int n) {
//   int amari = 0;
//   int keta = 0;
//   while (n > 0) { // n が 0 になるまで
//     amari += n % 2;
//     if (keta%2==0)
//     n /= 10;
//   }
//   return sum;
// }
 
ll gcd(ll a, ll b) {
	if(b == 0) return a;
	return gcd(b, a % b);
}

// const ll MAX_N = 1e5;
// ll N;
// ll h[MAX_N];
// ll dp[MAX_N+1]; // メモ化テーブル
 
// // i番目以降の品物から重さの総和がj以下になるように選ぶ
// void rec(ll i){
// 	if (i==0) dp[i] = 0;
// 	else if (i==1) dp[i] = abs(h[1]-h[0]);
// 	else dp[i] = min(dp[i-1]+abs(h[i]-h[i-1]), dp[i-2]+abs(h[i]-h[i-2]));
// 	return;
// }
 
// https://cf17-relay-open.contest.atcoder.jp/tasks/relay2_e
int main() {
	// 入力
	ll N;
	float Percent;
	cin >> N>>Percent;
	Percent *= 0.01;
	vector<P> wb;
	for (ll i = 0; i < N; ++i){
		ll w,b;
		cin>>w>>b;
		wb.pb(make_pair(w,b));
	}
 
	// 解法
	// REP(i,N) printf("%lld, %lld\n", wb[i].first, wb[i].second);
	float tmp_w = 0; // 白票合計
	float tmp_b = 0;
	REP(i,N) tmp_b += float(wb[i].second); // 全員青に入れた場合の票数
	while (tmp_w / (tmp_w + tmp_b)<Percent){
		// printf("%lld, %f, %f\n", N-wb.size(), tmp_w, tmp_b);
		// printf("%lld, %f\n", N-wb.size(), tmp_w/(tmp_w + tmp_b));
		float tmp = tmp_w / (tmp_w + tmp_b);
		ll tmp_idx = 0; // 寝返ると最も得票率を上げる議員のindex
		REP(i, wb.size()){
			float Percent_tmp = (tmp_w + float(wb[i].first))
							 / (tmp_w + tmp_b + float(wb[i].first - wb[i].second));
			if (tmp<Percent_tmp){
				tmp_idx = i;
				tmp = Percent_tmp;
			}
			// printf("%lld, %f %f\n", i, tmp, Percent_tmp);
		}
		tmp_w += wb[tmp_idx].first;
		tmp_b -= wb[tmp_idx].second;
		wb.erase (wb.begin()+tmp_idx); // 選択した議員を除去
	}
	ll ans = N - wb.size();
	
	//出力
	cout << ans << endl;
}