#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl "\n"
#define ull unsigned long long
#define ms(x, y) memset(x, y, sizeof x);
#define debug(x) cout << #x << " = " << x << endl;
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#define fre                           \
freopen("input.txt", "r", stdin); \
freopen("output.txt", "w", stdout);
const int mod = 998244353;
const int inf = 0x3f3f3f3f3f3f3f3f;
const int N = 1e6 + 10;
const double esp = 1e-6;
const ull MOD1 = 1610612741;
const ull MOD2 = 805306457;
const ull BASE1 = 1331;
const ull BASE2 = 131;
#define pre(i, a, b) for (int i = a; i <= b; i++)
#define rep(i, a, b) for (int i = a; i >= b; i--)
#define all(x) (x).begin(), (x).end()
char *p1, *p2, buf[100000]; // 快读和同步流二者只能选一个
#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++)
int read()
{
	int x = 0, f = 1;
	char ch = nc();
	while (ch < 48 || ch > 57)
	{
		if (ch == '-')
			f = -1;
		ch = nc();
	}
	while (ch >= 48 && ch <= 57)
		x = x * 10 + ch - 48, ch = nc();
	return x * f;
}
void write(int x)
{
	if (x < 0)
		putchar('-'), x = -x;
	if (x > 9)
		write(x / 10);
	putchar(x % 10 + '0');
	return;
}
//每轮之后 a[i] 都会加到 now 上,当前now > 当前最高评价时需要更新最高评价
//找出最小值和最大值即可,两者之间数据一定是可以到达的
//最大的一定是正数递增先排,答案就是正数个数
//最小一定是先负数,在正数递增,答案是前缀和大于 0 之后的后续位置
#define pa tr[u]
#define ls tr[u << 1]
#define rs tr[u << 1 | 1]
int n,m,a[N];
pair<int,int> q[N];
vector<int> ve;//离散化
//难度在于数据是变化的,需要动态维护,因为存在负数,树状数组处理可能会很麻烦,直接转换为权值线段树
struct node
{
    int x,sum;//当前权值左边的个数和区间和
}tr[N << 2];
node mer(node l,node r)
{
    node res = {0,0};
    res.x = l.x + r.x;
    res.sum = l.sum + r.sum;
    return res;
}
void build(int u,int l,int r)
{
    if(l > r) return;
    if(l == r)
    {
        pa.x = pa.sum = 0;
        return ;
    }
    int mid = l + r >> 1;
    build(u << 1,l,mid);
    build(u << 1,mid + 1,r);
    pa = mer(ls,rs);
}
void modify(int u,int l,int r,int L,int R) //单点修改,先删除后修改
{
    if(l > r) return ;
    if(L <= l && r <= R)
    {
        pa.x ++;
        pa.sum += ve[L - 1];
        // cout << L << " ccccc " << pa.x << " " << pa.sum << endl;
        return ;
    }
    int mid = l + r >> 1;
    if(L <= mid)
        modify(u << 1,l,mid,L,R);
    if(R > mid)
        modify(u << 1 | 1,mid + 1,r,L,R);
    pa = mer(ls,rs);
    return ;
}
void del(int u,int l,int r,int L,int R)
{
    if(l > r) return ;
    // cout << L << " dddddd " << R << endl;
    // cout << l << " " << r << " " << L << " " << R << endl;
    if(L <= l && r <= R)
    {
        pa.x --;
        pa.sum -= ve[L - 1];
        // cout << L << " ddddd " << pa.x << " " << pa.sum << endl;
        return ;
    }
    int mid = l + r >> 1;
    if(L <= mid)
        del(u << 1,l,mid,L,R);
    if(R > mid)
        del(u << 1 | 1,mid + 1,r,L,R);
    pa = mer(ls,rs);
    return ;
}
int query(int u,int l,int r,int x)//二分查找目前的最小值,最大值可以实时维护
{//寻找目前前缀和大于 0 的位置
    if(l > r) return -1;
    if(tr[1].sum <= x) return -1;//找不到
    if(l == r)
        return x / ve[l - 1] + 1;//此时是单点,判断这个点是整数是负数??
    int mid = l + r >> 1;
    if(ls.sum > x) return query(u << 1,l,mid,x);
    else return ls.x + query(u << 1 | 1,mid + 1,r,x - ls.sum);
}
int find(int x)
{
    return lower_bound(ve.begin(),ve.end(),x) - ve.begin()  + 1;
}
void solve()
{
	cin >> n >> m;
    int mx = 0,sum = 0;
    pre(i,1,n)
    {
        cin >> a[i];
        if(a[i] > 0) mx ++,ve.push_back(a[i]);
        else sum -= a[i];
    }
    pre(i,1,m)
    {
        cin >> q[i].first >> q[i].second;
        if(q[i].second > 0)
            ve.push_back(q[i].second);
    }
    sort(ve.begin(),ve.end());
    ve.erase(unique(ve.begin(),ve.end()),ve.end());
    int total = ve.size();
    // cout << total << endl;
    build(1,1,total);
    pre(i,1,n)
    {
        if(a[i] <= 0)
        {
            sum -= a[i];//负数直接使用sum统计即可
        }
        else //权值线段树只保留正数
        {
            a[i] = find(a[i]);//离散化到下标即可
            modify(1,1,total,a[i],a[i]);
        }
    }
    pre(i,1,m)
    {
        if(total == 0)
        {
            cout << 1 << endl;
            continue;
        }
        if(a[q[i].first] <= 0)
        {
            sum += a[q[i].first];
        }     
        else 
        {
            // cout << "this" << " " << q[i].first << " " << a[q[i].first] << endl;
            del(1,1,total,a[q[i].first],a[q[i].first]);
            mx -- ;
        }
        if(q[i].second <= 0)
            sum -= q[i].second ,a[q[i].first] = q[i].second;
        else 
        {
            a[q[i].first] = find(q[i].second);
            modify(1,1,total,a[q[i].first],a[q[i].first]);
            mx ++ ;
        }
        int res = query(1,1,total,sum);//需要多少个正数在前面
        // cout << i << " " << res << " " << tr[1].sum << endl;
        cout << mx - (res == -1 ? 0 : mx - res + 1) + 1 << endl;
    }
}
// #define LOCAL
signed main()
{
	ios
	// fre
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	auto start = std::chrono::high_resolution_clock::now();
#endif
	
	int t = 1;
	// cin >> t;
	while (t--)
		solve();
	
#ifdef LOCAL
	auto end = std::chrono::high_resolution_clock::now();
	cout << "Execution time: "
	<< std::chrono::duration_cast<std::chrono::milliseconds>(end - start).count()
	<< " ms" << '\n';
#endif
	return 0;
}