Problem: H --Bit Operation Game

Given a rooted tree with N vertices. The vertices are numbered from 0 to N − 1, and the 0th vertex represents the root. `T = 0` for the root, but for the other vertices

* `T = T & X;`
* `T = T & Y;`
* `T = T | X`
* `T = T | Y`
* `T = T ^ X`
* `T = T ^ Y`



One of the operations is written. Here, the operators &, |, ^ mean the bitwise operators and, or, xor, respectively.

Mr. A and Mr. B played the following games M times using this tree. The two start from the root and select the child vertices and proceed alternately, starting from Mr. A and reaching the leaves. The final T value when the operations written in the passed nodes are applied in the order of passing is the score. Mr. B wants to make the score as small as possible, and Mr. A wants to make it large. Given the X and Y values ​​of the M games, output the score for each game when the two make the best choice.

Constraints

* 1 ≤ N ≤ 100000
* 1 ≤ M ≤ 100000
* 0 ≤ X, Y <2 ^ {16}



Input Format

Input is given from standard input in the following format.


N M
o_1_1
o_2
...
o_ {N−1}
u_1 v_1
u_2 v_2
...
u_ {N−1} v_ {N−1}
X_1 Y_1
X_2 Y_2
...
X_M Y_M


On the first line, the number of vertices N of the tree and the integer M representing the number of games played are entered.
From the 2nd line to the Nth line, the operation written at the 1st ... N-1st vertex is input.
In addition, the numbers of the two vertices connected by each side are entered in the N-1 line.
Finally, the values ​​of X and Y in M ​​games are entered over M lines.


Output Format

Print the final T value for each game on the M line.

Sample Input 1


6 3
T = T | X
T = T | Y
T = T | Y
T = T ^ Y
T = T & X;
0 1
0 2
13
14
twenty five
5 6
3 5
0 0


Sample Output 1


Four
6
0


<image>
For X = 5, Y = 6, proceed to vertices 0-> 2-> 5, and T = 5 & 6 = 4.
If X = 3, Y = 5, then go to vertices 0-> 1-> 4, and T = 3 ^ 5 = 6.
If X = 0, Y = 0, T does not change from 0 no matter where you go





Example

Input

6 3
T=T|X
T=T|Y
T=T|Y
T=T^Y
T=T&X
0 1
0 2
1 3
1 4
2 5
5 6
3 5
0 0


Output

4
6
0

Solution (in CPP):

Solution:
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = (1e+18) + 7;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-6;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

ll mod_pow(ll a, ll n) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%mod;
		a = a * a%mod; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

//const int max_n = 1 << 22;
//modint fact[max_n], factinv[max_n];
//void init_f() {
//	fact[0] = modint(1);
//	for (int i = 0; i < max_n - 1; i++) {
//		fact[i + 1] = fact[i] * modint(i + 1);
//	}
//	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
//	for (int i = max_n - 2; i >= 0; i--) {
//		factinv[i] = factinv[i + 1] * modint(i + 1);
//	}
//}
//modint comb(int a, int b) {
//	if (a < 0 || b < 0 || a < b)return 0;
//	return fact[a] * factinv[b] * factinv[a - b];
//}

int n, m;

string sord = "^|&";
int ord[1 << 17];
string sb = "XY";
int b[1 << 17];

int calc(int c,int r, int o, int x, int y) {
	if (o)swap(x, y);
	if (r == 0)return c ^ x;
	else if (r == 1)return c | x;
	else return c & x;
}

vector<int> G[1 << 17];

int dfs(int c,int x, int y, int id, int fr, int t) {
	int res = -1;
	if (t)res = mod;
	for (int to : G[id]) {
		if (to == fr)continue;
		int nc = calc(c, ord[to], b[to], x, y);
		int nex = dfs(nc, x, y, to, id, t ^ 1);
		if (t)res = min(res, nex);
		else res = max(res, nex);
	}
	//cout << id << " ! " << res <<" "<<c<< endl;
	if (res < 0 || res == mod)return c;
	else return res;
}

void solve() {
	cin >> n >> m;
	rep1(i, n - 1) {
		string s; cin >> s;
		ord[i] = sord.find(s[3]);
		b[i] = sb.find(s[4]);
	}
	rep(i, n - 1) {
		int a, b; cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	//cout << dfs(0, 3, 5, 0, -1, 0) << endl;
	vector<int> v = { 0,1,2,3 };
	map<vector<int>, int> mp;
	vector<int> memo;
	int tmp = 0;
	while (true) {
		mp[v] = tmp;
		int x = 0, y = 0;
		rep(i, 4) {
			int k = (1 << i);
			if (v[i] % 2)x += k;
			if (v[i] / 2)y += k;
		}
		int c = dfs(0, x, y, 0, -1, 0);
		memo.push_back(c);
		tmp++;
		if (!next_permutation(all(v)))break;
	}

	function<int(int, int)> calcans = [&](int x, int y)->int {
		vector<int> z;
		vector<bool> used(4, false);
		per(i, 16) {
			int u = 0;
			if (x&(1 << i))u |= 1;
			if (y&(1 << i))u |= 2;
			if (used[u])continue;
			used[u] = true; z.push_back(u);
		}
		rep(i, 4)if (!used[i])z.push_back(i);
		reverse(z.begin(), z.end());
		int id = mp[z];
		int val = memo[id];
		vector<int> g(4);
		rep(i, 4) {
			if (val&(1 << i))g[z[i]] = 1;
			else g[z[i]] = 0;
		}
		int res = 0;
		rep(i, 16) {
			int u = 0;
			if (x&(1 << i))u |= 1;
			if (y&(1 << i))u |= 2;
			if (g[u])res |= (1 << i);
		}
		return res;
	};
	rep(aa, m) {
		int x, y; cin >> x >> y;
		int ans = calcans(x, y);
		cout << ans << endl;
	}
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(12);
	//init_f();
	//int t; cin >> t; rep(i, t)solve();
	solve();
	stop
		return 0;
}