#include <iostream>
#include <vector>
using namespace std;

vector<int> son[100001];
long long dp[100001][3];
bool have[100001];
bool root[100001];

const int mod = 1000000007;

int dp2(int u, int c, int f)
{
	if (have[u])
	{
		return dp[u][c];
	}

	have[u] = true;
	for (int i = 0; i < son[u].size(); i++)
	{
		if (son[u][i] == f)
		{
			continue;
		}
		dp[u][0] *= (dp2(son[u][i], 1, u) + dp2(son[u][i], 2, u)) % mod;
		dp[u][1] *= (dp2(son[u][i], 0, u) + dp2(son[u][i], 2, u)) % mod;
		dp[u][2] *= (dp2(son[u][i], 0, u) + dp2(son[u][i], 1, u)) % mod;
	}
	return dp[u][c];
}

int main()
{
	int n, k;
	cin >> n >> k;

	for (int i = 1; i <= n; i++)
	{
		dp[i][0] = 1;
		dp[i][1] = 1;
		dp[i][2] = 1;
	}

	for (int i = 1; i < n; i++)
	{
		int a, b;
		cin >> a >> b;
		son[a].push_back(b);
		son[b].push_back(a);
		root[b] = true;
	}
	for (int i = 0; i < k; i++)
	{
		int u, c;
		cin >> u >> c;
		c--;
		dp[u][0] = (c == 0 ? 1 : 0);
		dp[u][1] = (c == 1 ? 1 : 0);
		dp[u][2] = (c == 2 ? 1 : 0);
	}

	int r = 1;
	cout << dp2(r, 0, -1) + dp2(r, 1, -1) + dp2(r, 2, -1);
	return 0;
}

用的记搜，但第4组输出0