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非常像数字游戏，再写一遍做法当复习了。

容易转化为赋值为 \(0 / 1\) 序列，如果 \(a_i \in [l, r]\)，则设为 \(1\)，查询则为 \(\sum \frac{len(len + 1)}{2}\)。

这很莫队。加入是简单的：每个连续的 \(1\) 段左右端点分别存下右左端点。将一个位置改为 \(1\) 的话，就相当于是合并两段 \(1\) 段。但你会发现删除一个位置是困难的。

于是回滚莫队即可。复杂度根号。

这题良心的是全局查，这样避免了再序列分块多维护一车东西。

#include <bits/stdc++.h>
#define _for(i, a, b)  for (int i = (a); i <= (b); i ++ )
#define _all(i, a, b)  for (int i = (a); i >= (b); i -- )
#define PII pair<int, int>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ll long long
using namespace std;
namespace IO {
	const int LIM = 1e5;
	static char buf[LIM], * p1 = buf, * p2 = buf, obuf[LIM], * p3 = obuf, cc[20];
	#define gc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, LIM, stdin), p1 == p2) ? EOF : * p1 ++ )
	#define pc(x) (p3 - obuf < LIM) ? ( * p3 ++ = x) : (fwrite(obuf, p3 - obuf, 1, stdout), p3 = obuf, * p3 ++ = x)
	inline int rd() {
		int val = 0, sgn = 1; char ch = gc();
		for ( ; ch < '0' || ch > '9'; ch = gc())  if (ch == '-')  sgn = - 1;
		for ( ; ch >= '0' && ch <= '9'; ch = gc())  val = (val << 3) + (val << 1) + (ch ^ 48); return (val * sgn);
	} inline void wt(ll x, char c) {
		if ( ! x)  pc('0'); int len = 0;
		if (x < 0)  x = - x, pc('-');
		while (x)  cc[len ++ ] = x % 10 | '0', x /= 10;
		while (len -- )  pc(cc[len]); pc(c);
	}
} using namespace IO;
const int N = 1e6 + 5;
int n, q, V, B, cnt, qcnt, ed[N], L[N], R[N], a[N], bel[N]; ll res, ans[N]; PII val[N], updL[N], updR[N];
struct Node { int l, r, id; } qry[N];
inline void upd_l(int x, int k) { updL[ ++ cnt] = mp(x, L[x]), L[x] = k; }
inline void upd_r(int x, int k) { updR[cnt] = mp(x, R[x]), R[x] = k; }
inline ll calc(int x) { return 1ll * x * (x + 1) / 2; }
inline void add(int op, int x) {
	if ( ! op) {
		if (L[x - 1] && R[x + 1])  R[L[x - 1]] = R[x + 1], L[R[x + 1]] = L[x - 1], res += calc(R[x + 1] - L[x - 1] + 1) - calc(x - L[x - 1]) - calc(R[x + 1] - x);
		else if (L[x - 1])  R[L[x - 1]] = x, L[x] = L[x - 1], res += x - L[x - 1] + 1;
		else if (R[x + 1])  L[R[x + 1]] = x, R[x] = R[x + 1], res += R[x + 1] - x + 1;
		else  L[x] = R[x] = x, res ++ ;
	} else {
		if (L[x - 1] && R[x + 1])  upd_l(R[x + 1], L[x - 1]), upd_r(L[x - 1], R[x + 1]), res += calc(R[x + 1] - L[x - 1] + 1) - calc(x - L[x - 1]) - calc(R[x + 1] - x);
		else if (L[x - 1])  upd_l(x, L[x - 1]), upd_r(L[x - 1], x), res += x - L[x - 1] + 1;
		else if (R[x + 1])  upd_l(R[x + 1], x), upd_r(x, R[x + 1]), res += R[x + 1] - x + 1;
		else  upd_l(x, x), upd_r(x, x), res ++ ;
	}
} inline ll calc(int op, int l, int r) {
	ll tmp = res, t;
	_for (i, l, op ? ed[bel[l]] : r)  add(1, val[i].se);
	_all (i, cnt, 1)  L[updL[i].fi] = updL[i].se, R[updR[i].fi] = updR[i].se; cnt = 0, t = res, res = tmp; return t;
} inline void solve() {
	n = rd(), q = rd(), B = sqrt(n); int l, r, cur, j; res = qcnt = 0;
	_for (i, 1, n)  bel[i] = (i - 1) / B + 1;
	_for (i, 1, n)  a[i] = rd(), val[i] = mp(a[i], i); sort(val + 1, val + n + 1, [&] (PII u, PII v) { return u.fi < v.fi; });
	_for (i, 1, bel[n])  ed[i] = min(n, i * B);
	_for (i, 1, q) {
		l = rd(), r = rd(); int LL = 1, RR = n, mid, pos = n + 1;
		while (LL <= RR) {
			mid = (LL + RR) >> 1;
			if (val[mid].fi >= l)  pos = mid, RR = mid - 1;
			else  LL = mid + 1;
		} l = pos, pos = 0, LL = 1, RR = n;
		while (LL <= RR) {
			mid = (LL + RR) >> 1;
			if (val[mid].fi <= r)  pos = mid, LL = mid + 1;
			else  RR = mid - 1;
		} r = pos;
		if (l > r)  ans[i] = 0;
		else if (bel[l] == bel[r])  ans[i] = calc(0, l, r);
		else  qry[ ++ qcnt] = (Node){l, r, i};
	} sort(qry + 1, qry + qcnt + 1, [&] (Node u, Node v) { return bel[u.l] ^ bel[v.l] ? bel[u.l] < bel[v.l] : u.r < v.r; }), j = 1;
	_for (i, 1, bel[n]) {
		cur = ed[i];
		while (j <= qcnt && bel[qry[j].l] < i)  j ++ ;
		while (j <= qcnt && bel[qry[j].l] == i) { { while (cur < qry[j].r)  add(0, val[ ++ cur].se); } ans[qry[j].id] = calc(1, qry[j].l, qry[j].r), j ++ ; }
		_for (j, 1, n)  L[j] = R[j] = 0; res = 0;
	} _for (i, 1, q)  wt(ans[i], '\n');
}
int main() {
	ios :: sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int T = rd(); while (T -- )  solve();
	return fwrite(obuf, p3 - obuf, 1, stdout), 0;
}