/*

dp维护整体的概率， 每次相当于回退一格然后重新dp一格 

*/

#include<cstdio>

#include<algorithm>

#include<iostream>

#include<cstring>

#include<queue>

#define ll long long

#define M 202

#define mmp make_pair

using namespace std;

int read()

{

	int nm = 0, f = 1;

	char c = getchar();

	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';

	return nm * f;

}

const int mod = 1000000007;

int poww(int a, int b)

{

	int ans = 1, tmp = a;

	for(; b; b >>= 1, tmp = 1ll * tmp * tmp % mod) if(b & 1) ans = 1ll * ans * tmp % mod;

	return ans;

}

void add(int &x, int y)

{

	x += y;

	x -= x >= mod ? mod : 0;

	x += x < 0 ? mod : 0;

}

struct Note

{

	int a, g, p;

	bool operator < (const Note &b) const

	{

		return this->a < b.a;

	}

}note[M][M], sta[M * M];

int m[M], q[M], n, v[M], tp;

int f[M], g[M], d[M], ans[M];

void work(int x)

{

	if(x == 0)

	{

		for(int i = 0; i <= n; i++) g[i] = g[i + 1];

		return;

	}

	int y = (1 + mod - x), invx = poww(x, mod - 2);

	memset(d, 0, sizeof(d));

	for(int i = 0; i <= n; i++)

	{

		d[i] = 1ll * g[i] * invx % mod;

		add(g[i + 1], -1ll * d[i] * y % mod);

	}

	for(int i = 0; i <= n; i++) g[i] = d[i];

}

int main()

{

	n = read();

	int inv = poww(100 ,mod - 2);

	for(int i = 1; i <= n; i++)

	{

		m[i] = read();

		for(int j = 1; j <= m[i]; j++)

		{

			note[i][j].a = read(), note[i][j].g = 1ll * (100 - read()) * inv % mod, note[i][j].p = read();

			add(q[i], note[i][j].p);

			sta[++tp] = (Note) {note[i][j].a, i, j};

		}

		int inv = poww(q[i], mod - 2);

		for(int j = 1; j <= m[i]; j++) note[i][j].p = 1ll * note[i][j].p * inv % mod;

	}

	sort(sta + 1, sta + tp + 1);

	for(int i = 1; i <= n; i++) v[i] = read();

	g[n] = 1;

	for(int now = 1; now <= tp; now++)

	{

		int i = sta[now].g, j = sta[now].p;

		work(f[i]);

		for(int a = 0; a <= n; a++) add(ans[i], 1ll * note[i][j].p * v[a + 1] % mod * g[a] % mod * note[i][j].g % mod);

		add(f[i], note[i][j].p);

		for(int a = n; a >= 0; a--)

		{

			 g[a] = 1ll * g[a] * f[i] % mod;

			 if(a) add(g[a], 1ll * (1 + mod - f[i]) * g[a - 1] % mod);

		}

	}

	for(int i = 1; i <= n; i++) cout << ans[i] << "\n";

	return 0;

}