Problem:
Score : 200 points

Problem Statement

Takahashi received otoshidama (New Year's money gifts) from N of his relatives.

You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either JPY or BTC, and x_i and u_i represent the content of the otoshidama from the i-th relative.

For example, if x_1 = 10000 and u_1 = JPY, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = 0.10000000 and u_2 = BTC, the otoshidama from the second relative is 0.1 bitcoins.

If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?

Constraints

2 \leq N \leq 10

u_i = JPY or BTC.

If u_i = JPY, x_i is an integer such that 1 \leq x_i \leq 10^8.

If u_i = BTC, x_i is a decimal with 8 decimal digits, such that 0.00000001 \leq x_i \leq 100.00000000.

Input

Input is given from Standard Input in the following format:

N
x_1 u_1
x_2 u_2
:
x_N u_N

Output

If the gifts are worth Y yen in total, print the value Y (not necessarily an integer).

Output will be judged correct when the absolute or relative error from the judge's output is at most 10^{-5}.

Sample Input 1

2
10000 JPY
0.10000000 BTC

Sample Output 1

48000.0

The otoshidama from the first relative is 10000 yen. The otoshidama from the second relative is 0.1 bitcoins, which is worth 38000.0 yen if converted at the rate of 380000.0 JPY per 1.0 BTC. The sum of these is 48000.0 yen.

Outputs such as 48000 and 48000.1 will also be judged correct.

Sample Input 2

3
100000000 JPY
100.00000000 BTC
0.00000001 BTC

Sample Output 2

138000000.0038

In this case, outputs such as 138001000 and 1.38e8 will also be judged correct.

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

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>

#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v,n) reverse(v,v+n);
#define VREVERSE(v) reverse(v.begin(), v.end());
#define ll long long
#define pb(a) push_back(a)
#define INF 9999999999
#define m0(x) memset(x,0,sizeof(x))
#define fill(x,y) memset(x,y,sizeof(x))
#define print(x) cout<<x<<endl;
#define pe(x) cout<<x<<" ";
#define lb(v,n) lower_bound(v.begin(), v.end(), n);
//#define int long long


using namespace std;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }


int dy[4] = { 0,0,1,-1 };
int dx[4] = { 1,-1,0,0 };
int dxx[9] = { 0, 0,0,1,1,1,-1,-1,-1 };
int dyy[9] = { 0, 1,-1,0,1,-1,0,1,-1 };




ll gcd(ll x, ll y) {
	ll m = max(x, y), n = min(x, y);
	if (m%n == 0)return n;
	else return gcd(m%n, n);
}
ll lcm(ll x, ll y) {
	return x / gcd(x, y)*y;
}

long long power_mod(long long x, long long n, long long m) {
	if (n == 0)
		return 1;
	if (n % 2 == 0)
		return power_mod(x * x % m, n / 2, m);
	else
		return x * power_mod(x, n - 1, m) % m;
}

long long power(long long a, long long n) {//aのn乗を計算します。
	long long x = 1;
	while (n > 0) {//全てのbitが捨てられるまで。
		if (n & 1) {//1番右のbitが1のとき。
			x = x * a;
		}
		a = a * a;
		n >>= 1;//bit全体を右に1つシフトして一番右を捨てる。
	}
	return x;
}



long long nCr(long long n, long long r) {
	if (r > n / 2) r = n - r; // because C(n, r) == C(n, n - r)
	long long ans = 1;
	long long i;

	for (i = 1; i <= r; i++) {
		ans *= n - r + i;
		ans /= i;
	}

	return ans;
}



const int MAX = 510000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

bool arr[100000100];
vector<ll>sosuu;
void Eratosthenes() {
	ll N = 10000010;
	int c = 0;
	for (int i = 0; i < N; i++) {
		arr[i] = 1;
	}
	for (ll i = 2; i < sqrt(N); i++) {
		if (arr[i]) {
			for (ll j = 0; i * (j + 2) < N; j++) {
				arr[i *(j + 2)] = 0;
			}
		}
	}

	for (ll i = 2; i < N; i++) {
		if (arr[i]) {
			sosuu.pb(i);
			//cout << sosuu[c] << " ";
			c++;
		}
	}
	//cout << endl;
	//cout << c << endl;
}

ll stoL(string s) {
	ll n = s.length();
	ll ans = 0;
	for (int i = 0; i < n; i++) {
		ans += power(10, n - i - 1)*(ll)(s[i] - '0');
	}
	return ans;
}

typedef pair<int, int> P;


void getbit(ll x, int bits[]) {
	int cnt = 0;
	while (x > 0) {
		if (x % 2 == 0)bits[cnt] = 0;
		else bits[cnt] = 1;
		x /= 2;
		cnt++;
	}
}
void addbit(ll x, int bits[]) {
	int cnt = 0;
	while (x > 0) {
		if (x % 2 == 0)bits[cnt] += 0;
		else bits[cnt] += 1;
		x /= 2;
		cnt++;
	}
}


int main() {
	int N; cin >> N;
	double ans = 0;
	REP(i, N) {
		double x;
		string u;
		cin >> x;
		cin >> u;
		if (u == "JPY") {
			ans += x;
		}
		else ans += x * 380000.0;
	}
	print(ans);
}