description: It is the hard version of the problem. The difference is that in this version, there are nodes with already chosen colors.
  
  Theofanis is starving, and he wants to eat his favorite food, sheftalia. However, he should first finish his homework. Can you help him with this problem?
  
  You have a perfect binary tree of 2^k - 1 nodes — a binary tree where all vertices i from 1 to 2^{k - 1} - 1 have exactly two children: vertices 2i and 2i + 1. Vertices from 2^{k - 1} to 2^k - 1 don't have any children. You want to color its vertices with the 6 Rubik's cube colors (White, Green, Red, Blue, Orange and Yellow).
  
  Let's call a coloring good when all edges connect nodes with colors that are neighboring sides in the Rubik's cube.
  
  <image>| <image>  
  ---|---  
  A picture of Rubik's cube and its 2D map.
  
  More formally: 
  
    * a white node can not be neighboring with white and yellow nodes; 
    * a yellow node can not be neighboring with white and yellow nodes; 
    * a green node can not be neighboring with green and blue nodes; 
    * a blue node can not be neighboring with green and blue nodes; 
    * a red node can not be neighboring with red and orange nodes; 
    * an orange node can not be neighboring with red and orange nodes; 
  
  
  
  However, there are n special nodes in the tree, colors of which are already chosen.
  
  You want to calculate the number of the good colorings of the binary tree. Two colorings are considered different if at least one node is colored with a different color.
  
  The answer may be too large, so output the answer modulo 10^9+7.
  
  Input
  
  The first line contains the integers k (1 ≤ k ≤ 60) — the number of levels in the perfect binary tree you need to color.
  
  The second line contains the integer n (1 ≤ n ≤ min(2^k - 1, 2000)) — the number of nodes, colors of which are already chosen.
  
  The next n lines contains integer v (1 ≤ v ≤ 2^k - 1) and string s — the index of the node and the color of the node (s is one of the white, yellow, green, blue, red and orange).
  
  It is guaranteed that each node v appears in the input at most once.
  
  Output
  
  Print one integer — the number of the different colorings modulo 10^9+7.
  
  Examples
  
  Input
  
  
  3
  2
  5 orange
  2 white
  
  
  Output
  
  
  1024
  
  
  Input
  
  
  2
  2
  1 white
  2 white
  
  
  Output
  
  
  0
  
  
  Input
  
  
  10
  3
  1 blue
  4 red
  5 orange
  
  
  Output
  
  
  328925088
  
  Note
  
  In the picture below, you can see one of the correct colorings of the first test example.
  
  <image>
solution: #include <bits/stdc++.h>
  using namespace std;
  const long long mod = 1e9 + 7;
  int k, n;
  long long memo[61][6];
  long long dp(int n, int c) {
    long long &ret = memo[n][c];
    if (n == k - 1) return ret = 1;
    if (ret != -1) return ret;
    ret = 0;
    int cc = c + 3;
    cc %= 6;
    long long tmp = 0;
    for (int i = 0; i < 6; i++)
      if (i != c && i != cc) ret += dp(n + 1, i);
    for (int i = 0; i < 6; i++)
      if (i != c && i != cc) tmp += dp(n + 1, i);
    ret %= mod;
    tmp %= mod;
    ret *= tmp;
    ret %= mod;
    return ret;
  }
  int main() {
    cin >> k >> n;
    vector<map<long long, long long>> a(61);
    for (int i = 0; i < n; i++) {
      long long v, tc;
      cin >> v;
      string s;
      cin >> s;
      if (s == "white") tc = 0;
      if (s == "yellow") tc = 3;
      if (s == "green") tc = 1;
      if (s == "blue") tc = 4;
      if (s == "red") tc = 2;
      if (s == "orange") tc = 5;
      long long tmp = 2;
      for (long long i = 0; i <= 60; i++, tmp *= 2)
        if (tmp > v) {
          a[i].insert({v, tc});
          break;
        }
    }
    for (int i = 0; i <= 60; i++)
      for (int j = 0; j < 6; j++) memo[i][j] = -1;
    for (int i = 0; i < 6; i++) dp(0, i);
    for (int i = 0; i < 6; i++) memo[k][i] = 0;
    map<long long, vector<long long>> now;
    for (int i = 60; i >= 0; i--)
      if (!now.empty() || !a[i].empty()) {
        map<long long, vector<long long>> next;
        for (auto it = now.begin(); it != now.end(); it++)
          if (a[i].find(it->first / 2) == a[i].end()) a[i][it->first / 2] = -10;
        for (auto it = a[i].begin(); it != a[i].end(); it++) {
          vector<long long> l, r;
          if (now.find(it->first * 2) == now.end())
            for (int k = 0; k < 6; k++) l.push_back(memo[i + 1][k]);
          else {
            l = now[it->first * 2];
            now.erase(now.find(it->first * 2));
          }
          if (now.find(it->first * 2 + 1) == now.end())
            for (int k = 0; k < 6; k++) r.push_back(memo[i + 1][k]);
          else {
            r = now[it->first * 2 + 1];
            now.erase(now.find(it->first * 2 + 1));
          }
          vector<long long> res(6);
          for (int j = 0; j < 6; j++) {
            if (it->second != -10 && it->second != j) continue;
            long long lsum = 0, rsum = 0;
            int c = j, cc = c + 3;
            cc %= 6;
            for (int k = 0; k < 6; k++)
              if (k != c && k != cc) lsum += l[k], rsum += r[k];
            lsum %= mod;
            rsum %= mod;
            lsum *= rsum;
            lsum %= mod;
            if (i == k - 1) lsum = 1;
            res[j] = lsum;
          }
          next.insert({it->first, res});
        }
        swap(now, next);
      }
    long long ans = 0;
    for (int i = 0; i < 6; i++) ans += now[1][i];
    ans %= mod;
    cout << ans;
  }
  