#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 8e5+5;
int sum[maxn<<2], num[maxn], dp1[maxn], dp2[maxn], dp3[maxn], dp4[maxn], a[maxn], b[maxn];
int x[maxn], y[maxn], ans[maxn];
vector <pll> v[maxn];
void update(int p, int val, int l, int r, int rt) {if(l == r) {sum[rt] = max(sum[rt], val);return ;}int m = l+r>>1;if(p<=m)update(p, val, lson);elseupdate(p, val, rson);sum[rt] = max(sum[rt<<1],sum[rt<<1|1]);
}
int query(int L, int R, int l, int r, int rt) {if(L>R)return 0;if(L<=l&&R>=r) {return sum[rt];}int m = l+r>>1, ret = 0;if(L<=m)ret = query(L, R, lson);if(R>m)ret = max(ret, query(L, R, rson));return ret;
}
int main()
{int n, m, cnt = 0;cin>>n>>m;for(int i = 1; i<=n; i++) {scanf("%d", &a[i]);b[cnt++] = a[i];}for(int i = 0; i<m; i++) {scanf("%d%d", &x[i], &y[i]);b[cnt++] = y[i];v[x[i]].pb(mk(i, y[i]));}sort(b, b+cnt);cnt = unique(b, b+cnt)-b;for(int i = 1; i<=n; i++) {a[i] = lower_bound(b, b+cnt, a[i])-b+1;}for(int i = 1; i<=n; i++) {dp1[i] = query(1, a[i]-1, 1, cnt, 1)+1;for(int j = 0; j<v[i].size(); j++) {int tmp = v[i][j].fi;int tmpy = lower_bound(b, b+cnt, v[i][j].se)-b+1;dp3[tmp] = query(1, tmpy-1, 1, cnt, 1)+1;}update(a[i], dp1[i], 1, cnt, 1);}mem(sum);for(int i = n; i>=1; i--) {dp2[i] = query(a[i]+1, cnt, 1, cnt, 1)+1;for(int j = 0; j<v[i].size(); j++) {int tmp = v[i][j].fi;int tmpy = lower_bound(b, b+cnt, v[i][j].se)-b+1;dp4[tmp] = query(tmpy+1, cnt, 1, cnt, 1)+1;}update(a[i], dp2[i], 1, cnt, 1);}int maxx = 0;for(int i = 1; i<=n; i++) {maxx = max(dp1[i]+dp2[i]-1, maxx);      //求原数列lis
    }for(int i = 1; i<=n; i++) {if(dp1[i]+dp2[i]-1 == maxx) {num[dp1[i]]++;              //如果num[dp1[i]] == 1, 那么它就是关键节点
        }}for(int i = 0; i<m; i++) {int tmp = maxx;if(dp1[x[i]]+dp2[x[i]]-1 == maxx && num[dp1[x[i]]] == 1) {tmp--;}tmp = max(tmp, dp3[i]+dp4[i]-1);ans[i] = tmp;}for(int i = 0; i<m; i++) {printf("%d\n", ans[i]);}return 0;
}