Claire is a man-eater. She's a real man-eater. She's going around with dozens of guys. She's dating all the time. And one day she found some conflicts in her date schedule. D'oh!

So she needs to pick some dates and give the others up. The dates are set by hours like 13:00 to 15:00. She may have more than one date with a guy. For example, she can have dates with Adam from 10:00 to 12:00 and from 14:00 to 16:00 and with Bob from 12:00 to 13:00 and from 18:00 to 20:00. She can have these dates as long as there is no overlap of time. Time of traveling, time of make-up, trouble from love triangles, and the likes are not of her concern. Thus she can keep all the dates with Adam and Bob in the previous example. All dates are set between 6:00 and 22:00 on the same day.

She wants to get the maximum amount of satisfaction in total. Each guy gives her some satisfaction if he has all scheduled dates. Let's say, for example, Adam's satisfaction is 100 and Bob's satisfaction is 200. Then, since she can make it with both guys, she can get 300 in total. Your task is to write a program to satisfy her demand. Then she could spend a few hours with you... if you really want.



Input

The input consists of a sequence of datasets. Each dataset has the following format:

N
Guy1
...
GuyN


The first line of the input contains an integer N (1 ≤ N ≤ 100), the number of guys. Then there come the descriptions of guys. Each description is given in this format:

M L
S1 E1
...
SM EM


The first line contains two integers Mi (1 ≤ Mi ≤ 16) and Li (1 ≤ Li ≤ 100,000,000), the number of dates set for the guy and the satisfaction she would get from him respectively. Then M lines follow. The i-th line contains two integers Si and Ei (6 ≤ Si < Ei ≤ 22), the starting and ending time of the i-th date.

The end of input is indicated by N = 0.

Output

For each dataset, output in a line the maximum amount of satisfaction she can get.

Example

Input

2
2 100
10 12
14 16
2 200
12 13
18 20
4
1 100
6 22
1 1000
6 22
1 10000
6 22
1 100000
6 22
16
1 100000000
6 7
1 100000000
7 8
1 100000000
8 9
1 100000000
9 10
1 100000000
10 11
1 100000000
11 12
1 100000000
12 13
1 100000000
13 14
1 100000000
14 15
1 100000000
15 16
1 100000000
16 17
1 100000000
17 18
1 100000000
18 19
1 100000000
19 20
1 100000000
20 21
1 100000000
21 22
0


Output

300
100000
1600000000
#include <bits/stdc++.h>
//typedef
//-------------------------#include <bits/stdc++.h>

const double pi = 3.141592653589793238462643383279;

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s)
{
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x)
{
  ostringstream sout;
  sout << x;
  return sout.str();
}
inline int readInt()
{
  int x;
  scanf("%d", &x);
  return x;
}

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<long long, long long> PLL;
typedef pair<int, PII> TIII;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;

//container util

//------------------------------------------
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a) * (a))
#define EACH(i, c) for (typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define EXIST(s, e) ((s).find(e) != (s).end())
#define SORT(c) sort((c).begin(), (c).end())

//repetition
//------------------------------------------
#define FOR(i, s, n) for (int i = s; i < (int)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define MOD 1000000007

#define rep(i, a, b) for (int i = a; i < (b); ++i)
#define trav(a, x) for (auto &a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()

typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, int> pii;
typedef vector<int> vi;
const double EPS = 1E-8;

#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)

class UnionFind
{
public:
  vector<int> par;
  vector<int> siz;

  UnionFind(int sz_) : par(sz_), siz(sz_, 1)
  {
    for (ll i = 0; i < sz_; ++i)
      par[i] = i;
  }
  void init(int sz_)
  {
    par.resize(sz_);
    siz.assign(sz_, 1LL);
    for (ll i = 0; i < sz_; ++i)
      par[i] = i;
  }

  int root(int x)
  {
    while (par[x] != x)
    {
      x = par[x] = par[par[x]];
    }
    return x;
  }

  bool merge(int x, int y)
  {
    x = root(x);
    y = root(y);
    if (x == y)
      return false;
    if (siz[x] < siz[y])
      swap(x, y);
    siz[x] += siz[y];
    par[y] = x;
    return true;
  }

  bool issame(int x, int y)
  {
    return root(x) == root(y);
  }

  int size(int x)
  {
    return siz[root(x)];
  }
};

ll modPow(ll x, ll n, ll mod = MOD)
{
  ll res = 1;
  while (n)
  {
    if (n & 1)
      res = (res * x) % mod;

    res %= mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}

#define SIEVE_SIZE 5000000 + 10
bool sieve[SIEVE_SIZE];
void makeSieve()
{
  for (int i = 0; i < SIEVE_SIZE; ++i)
    sieve[i] = true;
  sieve[0] = sieve[1] = false;
  for (int i = 2; i * i < SIEVE_SIZE; ++i)
    if (sieve[i])
      for (int j = 2; i * j < SIEVE_SIZE; ++j)
        sieve[i * j] = false;
}

bool isprime(ll n)
{
  if (n == 0 || n == 1)
    return false;
  for (ll i = 2; i * i <= n; ++i)
    if (n % i == 0)
      return false;
  return true;
}

const int MAX = 2000010;
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;
}

long long extGCD(long long a, long long b, long long &x, long long &y)
{
  if (b == 0)
  {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extGCD(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
// 負の数にも対応した mod (a = -11 とかでも OK)
inline long long mod(long long a, long long m)
{
  return (a % m + m) % m;
}

// 逆元計算 (ここでは a と m が互いに素であることが必要)
long long modinv(long long a, long long m)
{
  long long x, y;
  extGCD(a, m, x, y);
  return mod(x, m); // 気持ち的には x % m だが、x が負かもしれないので
}
ll GCD(ll a, ll b)
{

  if (b == 0)
    return a;
  return GCD(b, a % b);
}

typedef vector<ll> vec;
typedef vector<vec> mat;

mat mul(mat &A, mat &B)
{
  mat C(A.size(), vec((int)B[0].size()));
  for (int i = 0; i < A.size(); ++i)
  {
    for (int k = 0; k < B.size(); ++k)
    {
      for (int j = 0; j < B[0].size(); ++j)
      {
        C[i][j] = (C[i][j] + A[i][k] % MOD * B[k][j] % MOD) % MOD;
      }
    }
  }
  return C;
}
mat matPow(mat A, ll n)
{
  mat B(A.size(), vec((int)A.size()));

  for (int i = 0; i < A.size(); ++i)
  {
    B[i][i] = 1;
  }

  while (n > 0)
  {
    if (n & 1)
      B = mul(B, A);
    A = mul(A, A);
    n >>= 1;
  }
  return B;
}

int N;
vector<ll> M(110), L(110);
ll dp[110][(1 << 17)];
vector<int> vp;
ll rec(int idx, int mask)
{
  if (mask == (1 << 17) - 1)
  {
    return 0;
  }
  if (idx == N)
  {
    return 0;
  }
  if (dp[idx][mask] != -1)
    return dp[idx][mask];

  //cout << idx << " " << mask << " " << vp[idx] << endl;
  ll mx = rec(idx + 1, mask);

  if ((vp[idx] & mask) == 0)
  {
    mx = max(mx, rec(idx + 1, (mask | vp[idx])) + L[idx]);
  }
  return dp[idx][mask] = mx;
}

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(5);

  while (cin >> N)
  {
    if (N == 0)
      break;

    vp = vector<int>(N);
    REP(i, N + 1)
    {
      REP(j, (1 << 17))
      {
        dp[i][j] = -1;
      }
    }
    REP(i, N)
    {
      cin >> M[i] >> L[i];
      int mask = 0;
      REP(j, M[i])
      {
        int s, e;
        cin >> s >> e;
        for (int k = s - 6; k < e - 6; k++)
          mask |= (1 << k);
      }
      vp[i] = mask;
    }

    cout << rec(0, 0) << endl;
  }
  return 0;
}

