#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()
#define dump(x) cout << #x << " = " << (x) << endl
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }

template<class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
	os<<"{";
	rep(i, v.size()) {
		if (i) os<<",";
		os<<v[i];
	}
	os<<"}";
	return os;
}

const int MN = 100010;

// 1-indexed, [1, n]
template<class T>
class BIT {
	T bit[MN];
	int n;

public:
	BIT() {
		n = MN;
		memset(bit, 0, sizeof(bit));
	}

	T sum(int i) {
		++i;
		T s = 0;
		while (i > 0) {
			s += bit[i];
			i -= i & -i;
		}
		return s;
	}

	void add(int i, T x) {
		++i;
		while (i < n) {
			bit[i] += x;
			i += i & -i;
		}
	}
};

const int SZ = 1 << 17;

struct segtree {
	vector<int> dat;

	void init() {
		dat.resize(SZ * 2);
		rep(i, SZ * 2) dat[i] = INT_MIN;
	}

	void update(int pos, int x) {
		pos += SZ - 1;
		dat[pos] = x;

		while (pos > 0) {
			pos = (pos - 1) / 2;
			dat[pos] = max(dat[pos * 2 + 1], dat[pos * 2 + 2]);
		}
	}

	int get(int a, int b, int k = 0, int l = 0, int r = SZ) {
		if (b <= l || r <= a) return INT_MIN;
		if (a <= l && r <= b) return dat[k];
		return max(get(a, b, k * 2 + 1, l, (l + r) / 2), get(a, b, k * 2 + 2, (l + r) / 2, r));
	}
} seg;


int main() {
	seg.init();
	BIT<int> T;

	int N, Q; cin >> N >> Q;
	vi a(N), tol(N);

	rep(i, N) {
		cin >> a[i];
		--a[i];
		int t = seg.get(a[i] + 1, N);
		if (t == INT_MIN) {
			tol[i] = -1;
		} else {
			tol[i] = t;
		}
		seg.update(a[i], i);
	}
	dump(tol);

	vector<pii> vals;
	rep(i, N) vals.eb(tol[i], i);
	sort(ALL(vals));
	int now = 0;

	vector<vector<pii>> query(N);

	rep(i, Q) {
		int t, l, r;
		cin >> t >> l >> r;
		--l; --r;
		query[l].eb(r, i);
	}

	vi ans(Q, -1);

	rep(l, N) {
		while (now < N) {
			int x = vals[now].fi, p = vals[now].se;
			if (x < l) {
				T.add(p, 1);
				now++;
			} else {
				break;
			}
		}
		for (auto pr : query[l]) {
			int id = pr.se, r = pr.fi;
			ans[id] = T.sum(r) - T.sum(l-1);
		}
	}

	rep(i, Q) {
		printf("%d\n", ans[i]);
	}

	return 0;
}