// 膜强无敌的老师： https://paste.ubuntu.com/p/jFChknqmCD/inline void upd(int &x, int y) { x < y && (x = y); }const int N = 100005;int n, m;
int a[N], cnt[N], Max[N], ans[N];vector<int> v;
vector<P> q[N];void Inite() {v.clear();cnt[0] = Max[0] = 0;for (int i = 1; i <= n; ++i) {cnt[i] = cnt[i - 1];Max[i] = Max[i - 1];if (a[i] > Max[i - 1]) cnt[i]++, Max[i] = a[i];}
}int main()
{//freopen("D:\\1.in", "r", stdin);//freopen("D:\\1.txt", "w", stdout);
BEGIN() {sc(n), sc(m);for (int i = 1; i <= n; ++i) sc(a[i]), q[i].clear();Inite();for (int i = 1; i <= m; ++i) {int pos, x;sc(pos), sc(x);q[pos].pb(P(x, i));}for (int i = n; i >= 1; --i) {for (auto &p : q[i]) {int pos, x;tie(x, pos) = p;int res = cnt[i - 1] + (x > Max[i - 1]);upd(x, Max[i - 1]);res += v.rend() - upper_bound(v.rbegin(), v.rend(), x);ans[pos] = res;}while(v.size() && v.back() <= a[i]) v.pp();v.pb(a[i]);}for (int i = 1; i <= m; ++i) pr(ans[i]);}return 0;
}

inline void upd(int &x, int y) { x < y && (x = y); }
/* 比较扯的是 N = 100005 会RE，OMG，就真的不能一发过吗？ */
const int N = 200005;int n, m, tot, mPos;
int a[4 * N], b[N], cnt[N], Max[N], d[N];map<int, int> id;void Inite() {mem(d, 0);cnt[0] = Max[0] = 0;for (int i = 1; i <= n; ++i) {cnt[i] = cnt[i - 1];Max[i] = Max[i - 1];if (b[i] > Max[i - 1]) cnt[i]++, Max[i] = b[i];}
}inline void Pushup(int root) {a[root] = max(a[lson], a[rson]);
}void Build(int l, int r, int root) {if (l == r) {sc(a[root]);id[root] = ++tot;b[tot] = a[root];return;}int mid = (l + r) >> 1;Build(l, mid, lson);Build(mid + 1, r, rson);Pushup(root);
}

/*写挂了的查询！① mPos可能存在，但if过后直接退出而查不到那个位置去，② 当找到mPos后可能在其它节点再次被更新，而且会T，因为查询的代价可能不再是O（log（n））void Query(int l, int r, int root, int L, int R, int x) { 　　 if (l > R || r < L || L > R) return;    if (L <= l && r <= R && l == r) {        mPos = id[root];        return;    }    int mid = (l + r) >> 1;    if (a[lson] > x) Query(l, mid, lson, L, R, x);    else if (a[rson] > x) Query(mid + 1, r, rson, L, R, x);}*/void Query(int l, int r, int root, int L, int R, int x) {if (l > R || r < L || L > R || mPos != -1) return;if (L <= l && r <= R && l == r) {mPos = id[root];return;}int mid = (l + r) >> 1;if (a[lson] > x) Query(l, mid, lson, L, R, x);if (a[rson] > x) Query(mid + 1, r, rson, L, R, x);
}int main()
{//freopen("D:\\1.in", "r", stdin);//freopen("D:\\1.txt", "w", stdout);
BEGIN() {sc(n), sc(m);tot = 0;Build(1, n, 1);Inite();for (int i = n; i; --i) {mPos = -1;Query(1, n, 1, i + 1, n, b[i]);if (mPos == -1) d[i] = 1;else d[i] = d[mPos] + 1;}//for (int i = 1; i <= n; ++i) cout << i << " " << d[i] << endl;for (int i = 1; i <= m; ++i) {int pos, x;sc(pos), sc(x);int res = cnt[pos - 1] + (x > Max[pos - 1]);upd(x, Max[pos - 1]);mPos = -1;Query(1, n, 1, pos + 1, n, x);if (mPos != -1) res += d[mPos];pr(res);}}return 0;
}