#include <bits/stdc++.h>
using namespace std;
struct fastmod {
  typedef unsigned long long u64;
  typedef __uint128_t u128;

  int m;
  u64 b;

  fastmod(int m) : m(m), b(((u128)1 << 64) / m) {}
  int operator ()(u64 a) {
    u64 q = ((u128)a * b) >> 64;
    int r = a - q * m;
    return r < m ? r : r - m;
  }
} Red(2);

const int N = 15,M = 520,Lim = 5000;
const int B = 5,C = 3,S = 1 << B; //L : 块数，B : 块长

int n,Q,P;
inline int Add(const int &a,const int &b) { return (a + b >= P) ? (a + b - P) : (a + b);}
inline int Sub(const int &a,const int &b) { return (a < b) ? (a - b + P) : (a - b);}
inline void Plus(int &a,const int &b) { a += b;if(a >= P) a -= P;}
inline void Minus(int &a,const int &b) { a -= b;if(a < 0) a += P;}
inline int qpow(int a,int b) { int res = 1;while(b) {if(b&1) res = 1ll * res * a % P;a = 1ll * a * a % P;b >>= 1;} return res;}
typedef std::vector<int> vint;

vint f[C][S][M + 5]; // 块数，集合，k
vint g[C][S]; // 无限制

int fac[Lim + 5],ifac[Lim + 5],inv[Lim + 5];
int Poly[Lim + 5][M + 5];
void initfac(int n)
{
	fac[0] = 1;
	for(int i = 1;i <= n;i++) fac[i] = Red(1ll * fac[i - 1] * i);
	inv[1] = 1;
	for(int i = 2;i <= n;i++) inv[i] = Red(1ll * inv[P % i] * (P - P / i));
	ifac[0] = 1;
	for(int i = 1;i <= n;i++) ifac[i] = Red(1ll * ifac[i - 1] * inv[i]);
}
// 点值范围：0-Lim
void init_Lagrange(int n) // 系数表示法只需要保留 520 项，但是点值需要保留 2600/1560 项
{
	for(int i = 0;i <= n;i++)
		for(int j = 0;j <= min(n,M);j++)
			Poly[i][j] = 0;
	static int f[Lim + 5],tmp[Lim + 5];
	initfac(n + 1);
	memset(f,0,sizeof f);
	f[0] = 1;
	for(int i = 1;i <= n;i++)
	{
		for(int j = min(n,M) + 1;j >= 1;j--)
			f[j] = Red(1ll * f[j] * (P - i)),Plus(f[j],f[j - 1]);
		f[0] = Red(1ll * f[0] * (P - i));
	}
	for(int j = 0;j <= min(n,M);j++)
	{
		Poly[0][j] = Red(1ll * f[j] * ifac[n]);
		if(n & 1) Poly[0][j] = P - Poly[0][j];
	} 
	for(int j = min(n,M);j;j--) f[j] = f[j - 1];
	f[0] = 0;
	for(int i = 1;i <= n;i++)
	{
		memcpy(tmp,f,sizeof f);
		for(int j = 0;j <= min(n,M);j++)
			tmp[j] = P - Red(1ll * tmp[j] * inv[i]),Minus(tmp[j + 1],tmp[j]);
		int res = Red(1ll * ifac[i] * ifac[n - i]);
		if((n - i) & 1) res = P - res;
		for(int j = 0;j <= min(n,M);j++)
			Poly[i][j] = Red(1ll * res * tmp[j]);
	}
}
inline int Solve(int S,int k,int m)
{
	vector<int> tmp(M * (C + 1) + 1,1);
	for(int p = 0;p <= (n - 1) / B;S >>= B,++p)
	{
		int now = S & ((1 << B) - 1);
		for(int x = 0;x <= M * (C + 1);++x)
			tmp[x] = Red(1ll * tmp[x] * f[p][now][k][x]);
	}
	long long ans = 0;
	for(int x = 0;x <= M * (C + 1);++x)
		ans += Red(1ll * tmp[x] * Poly[x][m]);
	return ans % P;
}
int Cnt[S];
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin >> n >> Q >> P;
	Red = fastmod(P);
	for(int i = 0;i < n;i++)
	{
		int a;
		cin >> a;
		static int t[M + 5];
		memset(t,0,sizeof t);
		t[0] = 1;
		for(int _ = 1;_ <= a;_++)
		{
			int b,c;
			cin >> b >> c;
			for(int j = b;j <= M;j++) Plus(t[j],t[j - b]);
			int tmp = b * (c + 1);
			if(tmp > M) continue;
			for(int j = M;j >= tmp;j--) Minus(t[j],t[j - tmp]);
		}
		// 没有限制的那个 g
		g[i / B][1 << (i % B)].resize(M * B + 1);
		for(int x = 0;x <= M * B;x++)
		{
			int res = 0;
			for(int j = M;j >= 0;j--)
				res = Red(1ll * res * x + t[j]);
			// printf("gres:%d\n",res)
			g[i / B][1 << (i % B)][x] = res;
		}
		for(int k = 0;k <= M;k++) f[i / B][1 << (i % B)][k].resize(M * (C + 1) + 1);
		for(int x = 0;x <= M * (C + 1);x++)
		{
			int powx = 1,res = 0;
			for(int k = 0;k <= M;k++)
			{
				res = Red(res + 1ll * t[k] * powx);
				f[i / B][1 << (i % B)][k][x] = P - res; // 容斥系数 -1
				powx = Red(1ll * powx * x);
			}
		}
	}
	for(int i = 1;i < S;i++) Cnt[i] = Cnt[i >> 1] + (i & 1);
	init_Lagrange(M * B); // 到时候要把点值个数由 M * B 变为 M * (C + 1) (由块内变为块间)
	for(int p = 0;p <= (n - 1) / B;++p)
	{
		int siz = min(B,n - p * B);
		g[p][0].assign(M * B + 1,1); // 空集的权都赋为 1
		for(int k = 0;k <= M;k++) f[p][0][k].assign(M * (C + 1) + 1,1);
		for(int S = 1;S < (1 << siz);++S)
			if(Cnt[S] > 1)
			{
				g[p][S].resize(M * B + 1);
				for(int x = 0;x <= M * B;++x)
					g[p][S][x] = Red(1ll * g[p][S & (S - 1)][x] * g[p][S & (-S)][x]);
				for(int k = 0;k <= M && (k + 1) * Cnt[S] <= M;k++)
				{
					f[p][S][k].resize(M * (C + 1) + 1);
					for(int x = 0;x <= M * (C + 1);++x)
						f[p][S][k][x] = Red(1ll * f[p][S & (S - 1)][k][x] * f[p][S & (-S)][k][x]);
				}
			}
		// return 0;
		// 只保留 M * (C + 1) + 1 个点值
		for(int S = 0;S < (1 << siz);++S)
		{
			vector<int> tmp(M + 1,0);
			for(int i = 0;i <= M;i++)
				for(int j = 0;j <= M * B;j++)
					Plus(tmp[i],Red(1ll * g[p][S][j] * Poly[j][i]));
			g[p][S].resize(M * (C + 1) + 1);
			for(int x = 0;x <= M * (C + 1);++x)
			{
				int res = 0;
				for(int j = M;j >= 0;j--)
					res = Red(1ll * res * x + tmp[j]);
				g[p][S][x] = res;
			}
		}
		// return 0;
		for(int S = 0;S < (1 << siz);++S)
			for(int k = 0;k <= M && (k + 1) * Cnt[S] <= M;k++)
				for(int x = 0;x <= M * (C + 1);++x)
					f[p][S][k][x] = Red(1ll * f[p][S][k][x] * g[p][((1 << siz) - 1) ^ S][x]);
		// return 0;
		for(int u = 0;u < siz;++u)
			for(int S = 0;S < (1 << siz);++S)
				if((S >> u) & 1)
					for(int k = 0;k <= M;k++)
						for(int x = 0;x <= M * (C + 1);++x)
							Plus(f[p][S][k][x],f[p][S ^ (1 << u)][k][x]);
	}
	init_Lagrange(M * (C + 1));
	for(int i = 0;i <= M * (C + 1);i++)
		for(int j = 1;j <= M;j++)
			Plus(Poly[i][j],Poly[i][j - 1]);
	while(Q--)
	{
		string s;
		int k,m;
		cin >> s >> m >> k;
		--k;
		int S = 0;
		for(int i = n - 1;i >= 0;i--)
			S = S << 1 | (s[i] - '0');
		cout << Solve(S,k,m) << endl;
	}
	return 0;
}