#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define vi vector<int>
#define pi pair<int,int>
#define sz(v) (int)(v).size()
#define all(v) (v).begin(), (v).end()
#define compact(v) (v).erase(unique(all(v)), (v).end())
template<class T> using upq = priority_queue<T, vector<T>, greater<T>>;
template<class T> int lwrbound(const vector<T>& a, const T& b, const int s = 0){return int(lower_bound(s + all(a), b) - a.begin());}
template<class T> int uprbound(const vector<T>& a, const T& b, const int s = 0){return int(upper_bound(s + all(a), b) - a.begin());}
#define FOR(i, a, b) for(int i = (a); i <= (b); i++)
#define ROF(i, a, b) for(int i = (a); i >= (b); i--)
#define sumof(x) accumulate(all(x), 0ll)
#define dbg(x) "[" << #x " = " << (x) << "]"
#define el "\n"
using ll = long long;
using ld = long double;
template<class T> bool ckmx(T& a, const T b){return (a < b ? a = b, true : false);}
template<class T> bool ckmn(T& a, const T b){return (a > b ? a = b, true : false);}
const int N = 2e5 + 5;
const int MOD = 1e9 + 7;
const ll INF = numeric_limits<ll>::max() / 8;
/* (i, j): state of col i vs row j
calculate the number that has (1, 0) or (0, 1) is hard
so we will calculate which has both (0,0) and (1,1)
*/
struct SMT{
vector<int> st, cnt; int trsz;
SMT(int n = 0): trsz(1 << n), st((1 << n) << 2 | 1), cnt(n) {
for(int i = 0; i < n; i++){
cnt[i] = (1 << i);
}
}
void update(int id, int l, int r, int x, int depth){
if(l == r){
st[id] ^= 1;
}
else{
int m = l+r>>1;
if(x <= m)
update(id << 1, l, m, x, depth + 1);
else update(id << 1|1, m+1, r, x, depth + 1);
assert(depth < sz(cnt));
if(st[id] < 2) --cnt[depth];
if(max(st[id << 1], st[id << 1|1]) > 1){
st[id] = 2;
}
else{
if(st[id << 1] == st[id << 1|1]){
st[id] = st[id << 1];
++cnt[depth];
}
else st[id] = 2;
}
}
}
void update(int x){
update(1, 1, trsz, x, 0);
}
ll query(int x){
return cnt[x];
}
};
void Main()
{
int n, q; cin >> n >> q;
vector<SMT> tr(2, SMT(n));
while(q--){
int t,x; cin >> t >> x;
tr[t].update(x);
ll ans = 0;
for(int i = 0; i < n; i++){
ans += (1ll << i) * (1ll << i);
ans -= tr[0].query(i) * tr[1].query(i);
}
// assume each point will become a square when (sum of compress of four photo) + 1
cout << ans * 4 + 1 << el;
}
}
int32_t main()
{
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
#define name "InvMOD"
if(fopen(name".INP", "r")){
freopen(name".INP", "r", stdin);
freopen(name".OUT", "w", stdout);
}
int t = 1; while(t--) Main();
return 0;
}
